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Question Number 154384 by mathdanisur last updated on 17/Sep/21

$$\mathrm{If}\mathrm{x};\mathrm{y};\mathrm{z}>0\mathrm{then}:$$$$\frac{{4\mathrm{x}}^2}{\mathrm{x}+\mathrm{y}}+\frac{{8\mathrm{y}}^2}{\mathrm{y}+\mathrm{z}}+\frac{{4\mathrm{z}}^2}{\mathrm{z}+\mathrm{x}}⩾2\mathrm{x}+5\mathrm{y}+\mathrm{z}$$

Answered by ghimisi last updated on 18/Sep/21

$$4(\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{y^2}{y+z}+\frac{z^2}{x+z})⩾4\cdot \frac{{(x+y+y+z)}^2}{x+y+y+z+y+z+x+z}=$$$$=4\cdot \frac{{(x+2y+z)}^2}{2x+3y+2z}\stackrel{\bullet }{⩾}2x+5y+z$$$$\bullet \iff 4{(x+2y+z)}^2⩾(2x+5y+z)(2x+3y+2z)\iff ...\iff {(y-z)}^2⩾0$$

Commented by mathdanisur last updated on 18/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathrm{Ser},\mathrm{thankyou}$$

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