Question

Let $x,y,z$ are positive real numbers and $l_1$ is the least value of $2x^4+2y^4+4z^4-8xyz$ and $l_2$ is the least value of $x^{4}y+x{y}^4+\frac{4}{x^2{y}^3}+\frac{1}{x^3{y}^2}+8$. Then

• A. $l_2>-1$
• B. $l_2=10$
• C. $l_1=-1$
• D. $l_2>10$

Answer 1

Answer: (B), (C)

$l_1=-1$

For least value of $2x^4+2y^4+4z^4-8xyz$
Using $A.M.$ $\ge G.M.$,
$\frac{2x^4+2y^4+4z^4+1}{4}\ge (2x^4\times 2y^4\times 4z^4\times 1{)}^{\frac{1}{4}}$
$\Rightarrow \frac{2x^4+2y^4+4z^4+1}{4}\ge 2\left(xyz\right)$
$\Rightarrow 2x^4+2y^4+4z^4+1\ge 8xyz$
$\Rightarrow 2x^4+2y^4+4z^4-8xyz\ge -1$
$\therefore l_1=-1$

Again $A.M.$ $\ge G.M.$
$\Rightarrow \frac{x^{4}y+x{y}^4+\frac{4}{x^2{y}^3}+\frac{1}{x^3{y}^2}+8}{5}\ge {(x^{4}y\times x{y}^4\times \frac{4}{x^2{y}^3}\times \frac{1}{x^3{y}^2}\times 8)}^{\frac{1}{5}}$
$\Rightarrow \frac{x^{4}y+x{y}^4+\frac{4}{x^2{y}^3}+\frac{1}{x^3{y}^2}+8}{5}\ge (2^5{)}^{\frac{1}{5}}$
$\Rightarrow x^{4}y+x{y}^4+\frac{4}{x^2{y}^3}+\frac{1}{x^3{y}^2}+8\ge 2\times 5$
$\Rightarrow x^{4}y+x{y}^4+\frac{4}{x^2{y}^3}+\frac{1}{x^3{y}^2}+8\ge 10.$