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Question Number 24469 by Tinkutara last updated on 18/Nov/17

Let 2x + 3y + 4z = 9, x, y, z > 0 then the maximum value of (1 + x)^2 (2 + y)^3 (4 + z)^4 is

$$\mathrm{Let}\:\mathrm{2}{x}\:+\:\mathrm{3}{y}\:+\:\mathrm{4}{z}\:=\:\mathrm{9},\:{x},\:{y},\:{z}\:>\:\mathrm{0}\:\mathrm{then} \\ $$ $$\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}\:+\:{x}\right)^{\mathrm{2}} \:\left(\mathrm{2}\:+\:{y}\right)^{\mathrm{3}} \\ $$ $$\left(\mathrm{4}\:+\:{z}\right)^{\mathrm{4}} \:\mathrm{is} \\ $$

Commented bysushmitak last updated on 18/Nov/17

AM>GM (((1+x)×2+(2+y)×3+(4+z)×4)/9)≥((1+x)^2 (2+y)^3 (4+z)^4 )^(1/9) ((24+9)/9)≥((1+x)^2 (2+y)^3 (4+z)^4 )^(1/9) ((11)/3)≥((1+x)^2 (2+y)^3 (4+z)^4 )^(1/9) (((11)/3))^9 ≥(1+x)^2 (2+y)^3 (4+z)^4

$$\mathrm{AM}>\mathrm{GM} \\ $$ $$\frac{\left(\mathrm{1}+{x}\right)×\mathrm{2}+\left(\mathrm{2}+{y}\right)×\mathrm{3}+\left(\mathrm{4}+{z}\right)×\mathrm{4}}{\mathrm{9}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$ $$\frac{\mathrm{24}+\mathrm{9}}{\mathrm{9}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$ $$\frac{\mathrm{11}}{\mathrm{3}}\geqslant\left(\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{9}} \\ $$ $$\left(\frac{\mathrm{11}}{\mathrm{3}}\right)^{\mathrm{9}} \geqslant\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{2}+{y}\right)^{\mathrm{3}} \left(\mathrm{4}+{z}\right)^{\mathrm{4}} \\ $$

Commented byTinkutara last updated on 18/Nov/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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