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Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$ prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$

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Two downvotes and a close vote ("too localized"?) within five minutes and not even a single constructive comment to communicate with the OP? Not cool. –  anon Aug 23 '12 at 11:31
    
agree with @anon, I don't see what's wrong with this question. Otherwise I'd suggest to study the function$\varphi(a,b,c)=\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt‌​{c+a}}$ and prove that it reaches its minimum when $a=b=c$. You shouldnt need to usethe other constraints. –  S4M Aug 23 '12 at 11:37
    
I think use arithmetic geometric to bound in terms of $\frac a {\sqrt{a+b}} \times $ etc. then concavity bounds $log(\frac{a+b} 2)\ge \frac 12 (log(a) + log(b))$ etc. to handle denominator –  mike Aug 23 '12 at 13:18
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Don: On this site we expect some motivation for questions, and we want to see that you have made some effort to answer the question yourself. Especially so in this case, because it looks like a homework problem. Click the "faq" link at the top of the page to read more about how to ask questions, homework in particular. –  Nate Eldredge Aug 24 '12 at 4:56
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"Motivation" ?? Huh ?? The motivation for many questions in pure mathematics is lost in the mists of time. If we need motivation, then how about (1) the inequaity has a nice symmetry to it, and (2) the proof is not immediately obvious to anyone here. –  bubba Aug 25 '12 at 3:08

3 Answers 3

Using Hölder's inequality we have:

$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2/3} (a(a+b)+b(b+c)+c(c+a))^{1/3}\geq a+b+c.$$

i.e.

$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2} \geq \frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca}.$$

We have to prove that:

$$\frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9}{2}.$$

i.e. $$2(a+b+c)^3\geq9\left(a^2+b^2+c^2+ab+bc+ca\right). \tag{1}$$

Let $p=a+b+c$ and $q=ab+bc+ca$ and using that $abc=1$ and $AM-GM$ we obtain that $q \geq 3$.

Inequality $(1)$ is equivalent to :

$$2p^3 \geq 9\left(p^2-2q+q\right) \Leftrightarrow 2p^3+9q \geq 9p^2.$$

Applying $AM-GM$ we obtain

$$2p^3+9q \geq 2p^3+27=p^3+p^3+27 \geq 3\cdot \sqrt[3]{27p^6}=9p^2,$$ as required.

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Nice proof (+1) –  Chris's sis Sep 11 '12 at 13:21

Let $a=b$. Then $c=\frac{1}{a^2}$ and the formula is : $$f(a)=\sqrt{\frac{a}{2}}+\sqrt{a+\frac{1}{a^2}}$$ $$f'(a)=\frac{1}{2\sqrt{2a}}+(\frac{1}{2}-\frac{1}{a^3}).\frac{1}{\sqrt{a+\frac{1}{a^2}}}$$

The only root in $[0,\infty)$ of f'(a) is 1, hence $f(1)=\frac{3}{\sqrt{2}}$ is a minimum.

Now, what happens if $a\neq b$ ? Use the same method, but say $b=k.a+(1-k)$ (so let $k=\frac{b-1}{a-1}$ be a constant, and if $a=1$ exchange $a$ and $c$). Then $c=\frac{1}{ka^2+a(1-k)}$ and :

$$f_k(a)=\frac{a}{\sqrt{a.(1+k)+(1-k)}}+\frac{ka+(1-k)}{\sqrt{ka+(1-k)+\frac{1}{ka^2+a(1-k)}}}+\frac{\frac{1}{ka^2+a(1-k)}}{\sqrt{a+\frac{1}{ka^2+a(1-k)}}}$$

Once again, obtain $f'_k$ and show that its only root is $1$ (This is quite technical, use mathematica ?). I agree there should have something more simple.

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The function $$f(x)=\frac{1}{\sqrt{x}}$$ is convex. Applying Jensen as follows : $$ a*f[a + b] + b*f[b + c] + c*f[c + a] >= (a + b + c)*f[\frac{ (a (a + b) + b (b + c) + c (c + a))}{(a + b + c)}] = \frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}} $$

We need to prove $$\frac{(a + b + c)^{3/2}}{\sqrt{a^2 + a b + b^2 + a c + b c + c^2}}>=\frac{3}{\sqrt{2}}$$

This is equivalent to proving:

$$2 (a + b + c)^3 - 9 (a^2 + a b + b^2 + a c + b c + c^2)>=0$$

Which is an easy exercise.

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