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Question Number 97485 by Farruxjano last updated on 08/Jun/20

Commented by prakash jain last updated on 09/Jun/20

$${\mathrm{E}}^2=$$$$x^{3}y+y^{3}z+z^{3}x+{xy}^3+{yz}^3+{zx}^3$$$$+2\sqrt{x^{3}y+y^{3}z+z^3}\sqrt{{xy}^2+{yz}^3+{zx}^3}$$$$⩽x^{3}y+y^{3}z+z^{3}x+{xy}^2+{yz}^3+{zx}^3$$$$+x^{3}y+y^{3}z+z^{3}x+{xy}^2+{yz}^3+{zx}^3$$$$=2(x^3(y+z)+y^3(z+x)+z^3(x+y))$$$$=2(x^3(2-x)+y^3(2-y)+z^3(2-z))$$$$=4(x^3+y^3+z^3)-2(x^4+y^4+z^4)$$$$\mathrm{Now}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{prove}$$$$4(x^3+y^3+z^3)-2(x^4+y^4+z^4)⩽4$$

Commented by Farruxjano last updated on 09/Jun/20

$$\mathrm{please},\mathrm{sir},\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{me}\mathrm{step}\mathrm{by}\mathrm{step}$$$$\mathrm{especially}\mathrm{here}:\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0\mathrm{and}\mathrm{so}...$$

Commented by prakash jain last updated on 09/Jun/20

$$\mathrm{That}\mathrm{is}\mathrm{what}\mathrm{i}\mathrm{got}\mathrm{in}\mathrm{step}2.$$$${\mathrm{E}}^2\mathrm{and}\mathrm{then}\mathrm{AM}-\mathrm{GM}$$

Answered by 1549442205 last updated on 17/Jun/20

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