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Blazing goIITian

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From AM-GM we have:

$x^{ rac{3}{2}}+y^{ rac{3}{2}}+y^ rac{3}{2}}geq 3sqrt{x}y$ and $x^{ rac{3}{2}}+z^{ rac{3}{2}}+z^ rac{3}{2}}geq 3sqrt{x}z$

So , $2({x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}})ge3sqrt{x}(y+z)$

or $rac{x}{y+z}geq rac{3}{2}cdot rac{x^{ rac{3}{2}}}{x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}}$

$sum rac{xsqrt{ x}}{y+z}geq rac{3}{2}cdotsum rac{x^{2}}{x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}}$

so we need to prove that $sum rac{x^{2}}{x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}}ge rac{1}{sqrt{3}}$

or $3(x^{2}+y^{2}+z^{2})^{2}geq (x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}})^{2}$

or $3(x^{2}+y^{2}+z^{2})^{2}(x+y+z)geq (x+y+z)(x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}})^{2}$ .............................[1]

From Cauchy–Schwarz inequality

$(x^{2}+y^{2}+z^{2})(x+y+z)geq (x^{ rac{3}{2}}+y^{ rac{3}{2}}+z^{ rac{3}{2}}})^{2}$ ......................[2]

and 3(x^2+y^2+z^2)ge(x+y+z) ...............................[3]

Multiplying [2] and [3] we get [1]. So it completes the proof ..

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