Let $a,b,c \ge 0 : ab+bc+ca=1.$ Prove that
$$\dfrac{1}{\sqrt{2a+bc}}+\dfrac{1}{\sqrt{2b+ca}}+\dfrac{1}{\sqrt{2c+ab}} \ge 1+\sqrt{2}.$$
Source: here.
and [here] If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$.
I tried Holder, or AM-GM, Cauchy – Schwarz but it is not good.
And this is an old Problem, it's easier.
Let $a,b,c \ge 0 : ab+bc+ca=1.$ Prove that
$$\dfrac{1}{2a+bc}+\dfrac{1}{2b+ca}+\dfrac{1}{2c+ab} \ge 2.$$
By AM-GM
$$\sum\limits_{cyc}\dfrac{1}{2a+bc} \ge \sum\limits_{cyc}\dfrac{1}{\dfrac{a}{b+c}+a(b+c)+ca} =\sum\limits_{cyc}\dfrac{b+c}{a+b+c}=2.$$
I hope this problem will be solved by a nice solution.
Thanks.
Best Answer
My second proof using Holder inequality.
By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}}\frac{1}{\sqrt{2a + bc}}\right)^2 \sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3\\ \ge{}& \left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3. \end{align*}
It suffices to prove that \begin{align*} &\left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3\\ \ge{}& (1 + \sqrt 2)^2\sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3. \tag{1} \end{align*}
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.
(1) is written as $$c_1 r + c_0 \ge 0\tag{2}$$ where \begin{align*} c_1 &= 42\,\sqrt {2}{p}^{2}+60\,\sqrt {2}p+56\,{p}^{2}-48\,\sqrt {2}+ 100\,p-88, \\ c_0 &= 10\,\sqrt {2}{p}^{3}-36\,\sqrt {2}{p}^{2}-8\,{p}^{ 3}+100\,\sqrt {2}p+92\,{p}^{2}-136\,\sqrt {2}-168\,p+32. \end{align*} We have $c_1 \ge 0.$
If $p^2 \ge 4q = 4$, we have $c_0 \ge 0$. Thus, (2) is true.
If $p^2 < 4q = 4$, using degree three Schur $r \ge \frac{4pq - p^3}{9} = \frac{4p - p^3}{9}$, we have $$c_1r + c_0 \ge c_1 \cdot \frac{4p - p^3}{9} + c_0 \ge 0.$$
We are done.