Two-Sided Inequality by Dorin Marghidanu

Solution 1

First off,

$\sqrt{2}(a+b+c)\ge c+\sqrt{2}(a+b)\ge c+\sqrt{2(a^2+b^2)},$

implying,

(1)

$\displaystyle \frac{1}{\sqrt{2}(a+b+c)}\le\frac{1}{c+\sqrt{2}(a^2+b^2)},$

with equality if $c=0 \text{ and } (a=0\text{ or }b=0).$ Also

$\sqrt{2(a^2+b^2)}\ge a+b\Rightarrow c+\sqrt{2(a^2+b^2)}\ge a+b+c,$

so that

(2)

$\displaystyle \frac{1}{c+\sqrt{2(a^2+b^2)}}\le\frac{1}{a+b+c},$

with equality if $a=b.$ From (1) and (2) we obtain

$\displaystyle \frac{1}{\sqrt{2}(a+b+c)}\le\frac{1}{c+\sqrt{2(a^2+b^2)}}\le\frac{1}{a+b+c}$

from which

$\displaystyle \frac{a+b}{\sqrt{2}(a+b+c)}\le\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\le\frac{a+b}{a+b+c}.$

$\displaystyle \sqrt{2}\le\sum_{cycl}\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\le 2.$

For the left inequality, equality holds if $(a,b,c)=(k,0,0),$ $k\gt 0,$ and permutations. For the right inequality, equality occurs if $a=b=c.$

Solution 2

By the AM-QM inequality,

\displaystyle\begin{align}&\frac{a+b}{c+\sqrt{2}(a^2+b^2)}\le\frac{a+b}{a+b+c}&\Rightarrow\\ &\sum_{cycl}\frac{a+b}{c+\sqrt{2}(a^2+b^2)}\le\sum_{cycl}\frac{a+b}{a+b+c}=2. \end{align}

For the left inequality, assume, WLOG, $a+b+c=1.$ We have $a^2+b^2\le (a+b)^2$ such that

\displaystyle \begin{align} &\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\ge\frac{a+b}{c+\sqrt{2}(a+b)}=\frac{a+b}{1+(\sqrt{2}-1)(a+b)}&\Rightarrow\\ &\sum_{cycl}\frac{a+b}{c+\sqrt{2(a^2+b^2)}}\ge \sum_{cycl}\frac{a+b}{1+(\sqrt{2}-1)(a+b)}. \end{align}

Observe now that the function $\displaystyle f(x)=\frac{x}{1+(\sqrt{2}-1)x}$ is concave on $[0,2]$ and the sequence $(1,1,0)$ majorizes $(a+b,b+c,c+a).$ Hence, by Karamata's inequality,

$\displaystyle \sum_{cycl}f(a+b)\ge f(1)+f(1)+f(0)=\sqrt{2}$

which completes the proof. On the right, equality holds for $a=b=c.$ On the left, for $(a,0,0)$ and permutations.

Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu, along with his solution (Solution 1). Solution 2 is by Leo Giugiuc.