Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$ prove that $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$
$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$
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0@bubba *Huh* yourself. Your depiction of *pure mathematics* is at best naive, but, to keep the discussion simple, one might recall, first, that to motivate questions is recommended by the site and, second, that Nate also mentioned a well-known *effort* criterion (which you do not comment). – 2012-08-25
3 Answers
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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0Nice proof (+1) – 2012-09-11
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.
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.