Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
New contributor
add a comment |
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
New contributor
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
I don't think thea.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
– Theo Bendit
Jan 3 at 23:59
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05
add a comment |
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
New contributor
Let x and a be real numbers > 0. Prove that $frac{1}{2}(x+frac{a}{x}) ge sqrt{a}$
My idea is that I'm going to use $a>b iff a^2>b^2$ since we are only dealing with postive real numbers we won't run into problems with the root.
$frac{1}{2}(x+frac{a}{x}) ge sqrt{a} iff (frac{1}{2}(x+frac{a}{x}))^2 = frac{1}{4}(x+frac{a}{x})(x+frac{a}{x})= frac{1}{4}(x^2+2a+frac{a^2}{x^2}) ge sqrt{a}^2=a$
And from that we get:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a$
I don't really know how to proceed from here. What about looking at different possibilities for the values of a and x?
Let x>a. Then:
$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge frac{1}{4}a^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a iff frac{1}{4}a^2x^2 +frac{1}{2}ax^2+frac{a^2}{4} ge ax^2$
And now I don't know how to prcoeed from here and doubt that my approach is correct. Can someone please help me out here? Tanks in advance.
inequality proof-writing
inequality proof-writing
New contributor
New contributor
edited Jan 3 at 23:59
Theo Bendit
16.7k12148
16.7k12148
New contributor
asked Jan 3 at 23:50
John D.
153
153
New contributor
New contributor
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
I don't think thea.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
– Theo Bendit
Jan 3 at 23:59
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05
add a comment |
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
I don't think thea.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.
– Theo Bendit
Jan 3 at 23:59
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
I don't think the
a.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.– Theo Bendit
Jan 3 at 23:59
I don't think the
a.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.– Theo Bendit
Jan 3 at 23:59
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05
add a comment |
3 Answers
3
active
oldest
votes
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
add a comment |
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
add a comment |
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
John D. is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061160%2fprove-that-frac12x-fracax-ge-sqrta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
add a comment |
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
add a comment |
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
From here:
$$frac{1}{4}x^2 +frac{1}{2}a+frac{a^2}{4x^2} ge a,$$
move the $a$ to the left side to yield
$$frac{1}{4}x^2 - frac{1}{2}a+frac{a^2}{4x^2} ge 0.$$
The left side is a perfect square:
$$left(frac{x}{2} - frac{a}{2x}right)^2.$$
answered Jan 3 at 23:53
Theo Bendit
16.7k12148
16.7k12148
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
add a comment |
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
Thank you. I was really hoping someone else saw the obvious, straightforward answer. :) +$1$
– Clayton
Jan 4 at 0:40
add a comment |
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
add a comment |
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
add a comment |
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
Because by AM-GM $$x+frac{a}{x}geq2sqrt{xcdotfrac{a}{x}}=2sqrt{a}.$$
answered Jan 3 at 23:54
Michael Rozenberg
97.2k1589188
97.2k1589188
add a comment |
add a comment |
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
add a comment |
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
add a comment |
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
$(sqrt{x}-sqrt{a/x})^2geq0Rightarrow x+a/x-2sqrt{a}geq 0Rightarrow 1/2(x+a/x)geq sqrt{a}.$
answered Jan 3 at 23:55
John_Wick
1,436111
1,436111
add a comment |
add a comment |
John D. is a new contributor. Be nice, and check out our Code of Conduct.
John D. is a new contributor. Be nice, and check out our Code of Conduct.
John D. is a new contributor. Be nice, and check out our Code of Conduct.
John D. is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061160%2fprove-that-frac12x-fracax-ge-sqrta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Use AM-GM inequality, like here
– rtybase
Jan 3 at 23:53
I don't think the
a.m.-g.m.-inequality
tag is necessarily appropriate here, as the asker shows no evidence of wanting to use it, or indeed knowing the inequality in the first place. This doesn't appear to be a question about the am-gm inequality.– Theo Bendit
Jan 3 at 23:59
I know the AM-GM inequality but am not allowed to use it
– John D.
Jan 4 at 0:00
@JohnD. if you know AM-GM inequality, then you should know how to prove it for 2 numbers. Apply the same strategy
– rtybase
Jan 4 at 0:05