Dan Sitaru's Cyclic Inequality In Three Variables with Constraints V

Problem

Dan  Sitaru's  Cyclic  Inequality In Three Variables with  Constraints V, problem

Solution

First of all, by the AM-GM inequality,

$\displaystyle \frac{1}{4uv}+\frac{1}{u+v}\ge \frac{2}{2\sqrt{uv(u+v)}}=\frac{1}{\sqrt{uv(u+v)}},$

with equality only if $4uv=u+v,$ i.e., $\displaystyle 4=\frac{1}{u}+\frac{1}{v}.$ It follows that

(1)

$\displaystyle \sum_{cycl}\frac{1}{\sqrt{ab(a+b)}}\le\frac{1}{4}\sum_{cycl}\frac{1}{ab}+\sum_{cycl}\frac{1}{a+b}.$

with equality only if $\displaystyle a=b=c=\frac{1}{2}.$

On the other hand,

$\displaystyle \begin{align} 3+\frac{a+b+c}{abc}&=\frac{ab+bc+ca}{2abc}+\frac{a+b+c}{4abc}\\ &=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right). \end{align}$

Thus, suffice it to prove that

(2)

$\displaystyle \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$

But $(u+v)^2\ge 4uv,$ implying $\displaystyle \frac{1}{4}\left(\frac{1}{u}+\frac{1}{v}\right)\ge \frac{1}{u+v},$ with equality only if $u=v.$ By adding, we obtain (2), with equality only if $a=b=c$ which, under the constraint, leads to $\displaystyle a=b=c=\frac{1}{2}.$

Extra

Let $a,b,c$ be positive real numbers, such that $ab+bc+ca=6abc.$ Prove that

$\displaystyle \frac{1}{\sqrt{ab(a+b)}}+\frac{1}{\sqrt{bc(b+c)}}+\frac{1}{\sqrt{ca(c+a)}}\le 6.$

The latter is equivalent to

$\displaystyle \frac{xy}{\sqrt{x+y}}+\frac{yz}{\sqrt{y+z}}+\frac{zx}{\sqrt{z+x}}\le\frac{3}{\sqrt{2}}.$

By the AM-GM inequality, $\sqrt{x+y}\ge\sqrt{2}\cdot\sqrt[4]{xy},$ etc. Thus, suffice it to show that $\displaystyle \frac{1}{3}\sum_{cycl}(xy)^{\frac{3}{4}}\le 1.$

By Jensen's inequality, $\displaystyle \frac{1}{3}\sum_{cycl}(xy)^{\frac{3}{4}}\le\left(\frac{xy+yz+zx}{3}\right)^{\frac{3}{4}}.$ But $\displaystyle \left(\frac{xy+yz+zx}{3}\right)^{\frac{3}{4}}\le 1,$ with equality at $x=y=z=1,$ i.e. $\displaystyle a=b=c=\frac{1}{2}.$

Acknowledgment

The problem was kindly posted by Dan Sitaru at the CutTheKnotMath facebook page, with several solutions. The above solution is by Theodoros Sampas; more can be found at the link.

I am grateful to Leo Giugiuc who noticed and pointed out a typo in the original formulation and also added the Extra observation.

 

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