Question-97485

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

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

E^2 = x^3 y+y^3 z+z^3 x+xy^3 +yz^3 +zx^3 +2(√(x^3 y+y^3 z+z^3 ))(√(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 ) Now we need to prove 4(x^3 +y^3 +z^3 )−2(x^4 +y^4 +z^4 ) ≤4

$$\mathrm{E}^{\mathrm{2}} = \\ $$$${x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {x}+{xy}^{\mathrm{3}} +{yz}^{\mathrm{3}} +{zx}^{\mathrm{3}} \\ $$$$+\mathrm{2}\sqrt{{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} }\sqrt{{xy}^{\mathrm{2}} +{yz}^{\mathrm{3}} +{zx}^{\mathrm{3}} } \\ $$$$\leqslant{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {x}+{xy}^{\mathrm{2}} +{yz}^{\mathrm{3}} +{zx}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:+{x}^{\mathrm{3}} {y}+{y}^{\mathrm{3}} {z}+{z}^{\mathrm{3}} {x}+{xy}^{\mathrm{2}} +{yz}^{\mathrm{3}} +{zx}^{\mathrm{3}} \\ $$$$=\mathrm{2}\left({x}^{\mathrm{3}} \left({y}+{z}\right)+{y}^{\mathrm{3}} \left({z}+{x}\right)+{z}^{\mathrm{3}} \left({x}+{y}\right)\right) \\ $$$$=\mathrm{2}\left({x}^{\mathrm{3}} \left(\mathrm{2}−{x}\right)+{y}^{\mathrm{3}} \left(\mathrm{2}−{y}\right)+{z}^{\mathrm{3}} \left(\mathrm{2}−{z}\right)\right) \\ $$$$=\mathrm{4}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)−\mathrm{2}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \right) \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{prove} \\ $$$$\mathrm{4}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{3}} \right)−\mathrm{2}\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \right)\:\leqslant\mathrm{4} \\ $$

Commented by Farruxjano last updated on 09/Jun/20

please, sir, can you explain me step by step especially here: (∂f/∂x)=0,(∂f/∂y)=0 and so...

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

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

That is what i got in step 2. E^2 and then AM−GM

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

Answered by 1549442205 last updated on 17/Jun/20

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