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# Which inequality is needed?

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

Note by Archit Boobna
2 years, 9 months ago

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

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??

- 2 years, 9 months 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

- 2 years, 9 months ago

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

Staff - 2 years, 9 months ago

Thanks!

- 2 years, 9 months ago

Thanks so much, its an amazing wiki

- 2 years, 9 months ago