Inequality-2

Relevant wiki: Cauchy-Schwarz Inequality

Note that $x+y+z=1$ $\implies x+y = 1-z$ , $y+z = 1-x$ and $z+x=1-y$ . Using Hölder's inequality as follows:

$\begin{aligned} \left(\frac x{\sqrt{1-x}} + \frac y{\sqrt{1-y}} + \frac z{\sqrt{1-z}}\right)^\frac 13 \left(\frac x{\sqrt{1-x}} + \frac y{\sqrt{1-y}} + \frac z{\sqrt{1-z}}\right)^\frac 13 \bigg(x(1-x) + y(1-y) + z(1-z) \bigg)^\frac 13 & \ge x + y + z = 1 \end{aligned}$

Cubing both sides:

$\begin{aligned} \left(\frac x{\sqrt{1-x}} + \frac y{\sqrt{1-y}} + \frac z{\sqrt{1-z}}\right)^2 \bigg(x(1-x) + y(1-y) + z(1-z) \bigg) & \ge 1 \end{aligned}$

$\begin{aligned} \implies \left(\frac x{\sqrt{1-x}} + \frac y{\sqrt{1-y}} + \frac z{\sqrt{1-z}}\right)^2 & \ge \frac 1{x+y+z-\color{#3D99F6}(x^2+y^2+z^2)} & \small \color{#3D99F6} \text{By Cauchy-Schwartz inequity:} \\ & \ge \frac 1{1-\color{#3D99F6}\frac 13} = \frac 32 & \small \color{#3D99F6} (x+y+z)^2 \le 3(x^2+y^2+z^2) \implies x^2+y^2+z^2 \ge \frac 13 \\ \implies \frac x{\sqrt{y+z}} + \frac y{\sqrt{z+x}} + \frac z{\sqrt{x+y}} & \ge \sqrt{\frac 32} & \small \color{#3D99F6} \text{Equality occurs when }x=y=z =\frac 13 \end{aligned}$

Therefore, $A+B = 3+2 = \boxed 5$ .

Let $a_1 = \frac{x}{\sqrt{z + y}}$ , $a_2 = \frac{y}{\sqrt{x + z}}$ , and $a_3 = \frac{z}{\sqrt{x + y}}$ .

Then by the AM-GM inequality , $\frac{a_1 + a_2 + a_3}{3} \geq \sqrt[3]{a_1a_2a_3}$ , and $a_1 + a_2 + a_3$ has a minimum value when $a_1 = a_2 = a_3$ . Therefore, $\frac{x}{\sqrt{z + y}} = \frac{y}{\sqrt{x + z}} = \frac{z}{\sqrt{x + y}}$ .

Since $x + y + z = 1$ , $z + y = x - 1$ , $x + z = y - 1$ , and $x + y = z - 1$ . Substituting these in the above equation, we have a minimum value when $\frac{x}{\sqrt{x - 1}} = \frac{y}{\sqrt{y - 1}} = \frac{z}{\sqrt{z - 1}}$ , so $x = y = z$ .

Since $x = y = z$ and $x + y + z = 1$ , $x = y = z = \frac{1}{3}$ , and the minimum value is $\frac{x}{\sqrt{z + y}} + \frac{y}{\sqrt{x + z}} + \frac{z}{\sqrt{x + y}} = \frac{\frac{1}{3}}{\sqrt{\frac{1}{3} + \frac{1}{3}}} + \frac{\frac{1}{3}}{\sqrt{\frac{1}{3} + \frac{1}{3}}} + \frac{\frac{1}{3}}{\sqrt{\frac{1}{3} + \frac{1}{3}}} = \sqrt{\frac{3}{2}}$ .

Therefore, $A = 3$ and $B = 2$ , and $A + B = \boxed{5}$ .

Relevant wiki: Jensen's Inequality

First solution :

Let $f(x)=\frac{x}{\sqrt{1-x}}, f’’(x)>0.$

Therefore, $\sum\frac{x}{\sqrt{1-x}} \ge \frac{3s}{\sqrt{1-s}}, \text { where } s=\frac{x+y+z}{3}=\frac{1}{3}$ .

By Jensen's inequality: $\frac{x}{\sqrt{z+y}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}} \ge \sqrt{ \dfrac 32}.$

Second Solution :

$\frac{x}{\sqrt{z+y}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}} \ge \frac{ 1}{3}(x+y+z)(\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}} +\frac{1}{\sqrt{x+y}})$

We know that $f{(x)}=\frac{1}{\sqrt{x}}$ is a convex function. so by Jensen's inequality:

$\frac{x}{\sqrt{z+y}}+\frac{y}{\sqrt{x+z}}+\frac{z}{\sqrt{x+y}}\ge \frac{1}{3}\Big(3\times\frac{1}{\sqrt{\frac{2}{3}(x+y+z)}}\Big)=\sqrt{\dfrac32.}$