Which inequality is needed?

If x+y+z=9, where x,y,z are positive reals, maximize ${x}^{4}{y}^{3}{z}^{2}$

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

$\frac{4x}{4}+\frac{3y}{3}+\frac{2z}{2}=9$

Applying concept of weighted means

$\huge{\frac{4\times \frac{x}{4}+3\times \frac{y}{3}+2\times \frac{z}{2}}{4+3+2}\geq [(\frac{x}{4})^{4}\times (\frac{y}{3})^{3}\times (\frac{z}{2})^{2}]^{\frac{1}{4+3+2}}}$

$\large{\frac{x+y+z}{9} \geq [\frac{x^4}{4^4}\times \frac{y^{3}}{3^3}\times \frac{z^2}{2^2}]^{\frac{1}{9}}}$

$1\geq [\frac{x^4}{4^4}\times \frac{y^{3}}{3^3}\times \frac{z^2}{2^2}]^{\frac{1}{9}}$

$\huge{\boxed{x^{4}y^{3}z^{2}\leq 4^4 \times 3^3 \times 2^2}}$

Is this the answer??

Tanishq Varshney - 6 years, 1 month ago

This is very Elementary and standard Questions ... It is twisted application of AM-GM ... Known as Weighted AM-GM inequality .

You Can Learn This from Here , wiki Page Airthmetic mean and geometric mean inequality

Karan Shekhawat - 6 years, 1 month ago

The other page that I would suggest is Applying AM-GM, which contains detailed explanations and scenarios for further applications.

Calvin Lin Staff - 6 years, 1 month ago

Thanks!

Archit Boobna - 6 years, 1 month ago

Thanks so much, its an amazing wiki

Archit Boobna - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$