## SUBJECT $2(7 p=4 p+3 p)$

a) Determine the minimum value of the expression $E(x, y)=x^{4}+y^{4}+\frac{2}{x^{2} y^{2}}, x, y \in \mathbb{R}$. Specify the values of $x$ and $y$ for which this minimum is achieved.

b) Determine the real numbers $x, y, z$ knowing that: $x+y+z=\frac{3}{2}$ and $x^{2}+y^{2}+z^{2}=\frac{3}{4}$.

(G.M. nr.10/2015)
Solution: a) We apply the (MA-MG) inequality twice

$$
E(x)=x^{4}+y^{4}+\frac{2}{x^{2} y^{2}} \geq 2 \sqrt{x^{4} y^{4}}+\frac{2}{x^{2} y^{2}}=2\left(x^{2} y^{2}+\frac{1}{x^{2} y^{2}}\right) \geq 2 \cdot 2 \cdot \sqrt{x^{2} y^{2} \cdot \frac{1}{x^{2} y^{2}}}=4
$$

## LOCAL STAGE MATHEMATICS OLYMPIAD - GRADE VIII

Both inequalities must become equalities, and this happens if on one hand $x^{4}=y^{4} \Leftrightarrow x= \pm y$, and on the other hand $x^{2} y^{2}=\frac{1}{x^{2} y^{2}} \Leftrightarrow(x y)^{4}=1 \Leftrightarrow x y= \pm 1$.

The solutions are immediately obtained: $(x, y) \in\{(1,1),(-1,1),(1,-1),(-1,-1)\}$

b) We make the substitutions $x-\frac{1}{2}=a, y-\frac{1}{2}=b, z-\frac{1}{2}=c, a, b, c \in \mathbb{R}$.

We have $a+b+c=\left(x-\frac{1}{2}\right)+\left(y-\frac{1}{2}\right)+\left(z-\frac{1}{2}\right)=(x+y+z)-\frac{3}{2}=0\left(^{*}\right)$

The relation $x^{2}+y^{2}+z^{2}=\frac{3}{4}$ becomes: $\left(a^{2}+a+\frac{1}{4}\right)+\left(b^{2}+b+\frac{1}{4}\right)+\left(c^{2}+c+\frac{1}{4}\right)=\frac{3}{4}$, from which, taking into account $\left(^{*}\right): a^{2}+b^{2}+c^{2}=0 \Rightarrow a=b=c$.

Therefore: $x=y=z=\frac{1}{2} \ldots \ldots .$.