Just another Inequality Problem

Let $\dfrac{x}{1-x^2}+\dfrac{y}{1-y^2}+\dfrac{z}{1-z^2} = M$

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By RMS inequality ,

$\dfrac{(x + y)^{2}}{2} \leq x^{2} + y^{2} = 1 - z^{2}$

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therefore

$\dfrac{z}{1 - z^{2}} \geq \dfrac{2z}{(x + y)^{2}}$

Thus ( for other 2 terms also)

$M \geq \dfrac{2x}{(z + y)^{2}} + \dfrac{2y}{(x + z)^{2}} + \dfrac{2z}{(x + y)^{2}}$ $... 1.$

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By Cauchy - Schwartz inequality,

$(x + y + z)( \dfrac{z}{(x + y)^{2}} + \dfrac{y}{(x + z)^{2}} + \dfrac{x}{(z + y)^{2}}) \geq ( \dfrac{z}{(x + y)} + \dfrac{y}{(x + z)} + \dfrac{x}{(z + y)})^{2}$ $...... 2.$

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$\dfrac{x}{(z + y)} + \dfrac{y}{(x + z)} + \dfrac{z}{(x + y)} \geq \frac{3}{2}$ $...3$ (can be easily proved as its symmetric in a,b,c)

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By RMS inequality,

$(\dfrac{x + y + z}{3})^{2} \leq \dfrac{x^{2} + y^{2} + z^{2}}{3} = \frac{1}{3}$ $... 4$

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For 1 by 2,3

$M(E) \geq \dfrac{2}{x + y + z}( \dfrac{z}{(x + y)} + \dfrac{y}{(x + z)} + \dfrac{x}{(z + y)})^{2}$

$E \geq \frac{2}{\sqrt{3}}(\frac{3}{2})^{2} = \frac{3\sqrt{3}}{2}$

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Let : $y\quad =\quad f\left( x \right) \quad =\quad \frac { \sqrt { x } }{ 1-x } \quad in\quad (0,1)$ .

Let us consider Three Points on the curve which forms an convex polygon (Here it is Triangle) Since Graph is concave up , And Centroid G of triangle must lie inside the triangle . Now drop Perpendicular from G point which meets the given curve at P point So x-coordinate of both are same !

${ y }_{ G }\quad \ge \quad { y }_{ p }\\ \\ \frac { \sum { f\left( { { x }_{ i } }^{ 2 } \right) } }{ n } \quad \ge \quad f\left( \cfrac { \sum { { { x }_{ i } }^{ 2 } } }{ n } \right) \\ \\ \sum { f\left( { { x }_{ i } }^{ 2 } \right) } \quad \ge \quad nf\left( \cfrac { 1 }{ n } \right) \quad \quad \quad (put\quad n\quad =\quad 3)\\ \\ \boxed { \sum { f\left( { { x }_{ i } }^{ 2 } \right) } \quad \ge \quad 3\cfrac { \sqrt { 3 } }{ 2 } } \quad Ans.$ .

Note that without loss of optimality, we can restrict the variables to the set $0\leq x,y,z \leq 1$ . Straightforward calculus gives that each of the function $\frac{x}{1-x^2}, \frac{y}{1-y^2}$ and $\frac{z}{1-z^2}$ is convex in that restricted set. Hence their sum is convex and also the constraint set $x^2+y^2+z^2=1$ is convex. Thus using KKT conditions of convex optimization theory, we note that the minimum value of the objective is attained at $x^*=y^*=z^*=\frac{1}{\sqrt{3}}$ . Hence $E=\frac{3\sqrt{3}}{2}$ .