<s>[INST] Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$ Let $x>1,y>1$ and $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$, then find the maximum value of $x^{\log_e y}$
My attempt is as follows:
As $x>1,y>1$ , so $\log_e x>0, \log_e y>0$, hence we can apply $AM>=GM$
$$\dfrac{log_e x+log_e y}{2}>=\sqrt{\log_e x\log_e y}$$
As both the sides are positive, so we can square both the sides without breaking the inequality.
$$(\log_e x)^2+(\log_e y)^2+2\log_e x\log_e y>=4\log_e x\log_e y$$
Using the given condition $\left(\log_e x \right)^2+\left(\log_e y \right)^2=\log(x^2)+log(y^2)$
$$(\log_e x^2)+(\log_e y^2)>=2\log_e x\log_e y$$
$$(\log_e x)+(\log_e y)>=\log_e x\log_e y$$
$$\log_e xy>=\log_e x^{\log_e y}$$
As $e>1$, so we can safely write $xy>=x^{\log_e y}$
But actual answer is $e^4$, I am not able to think of any other way. Please help me in this.
 [/INST] Let $\ln{x}=a$ and $\ln{y}=b$.
Thus, $a$ and $b$ are positives,  $$a^2+b^2=2(a+b)$$ and by C-S
$$2(a+b)=\frac{1}{2}(1+1)(a^2+b^2)\geq\frac{1}{2}(a+b)^2.$$
Thus, $$a+b\leq4$$ and by AM-GM:
$$2\sqrt{ab}\leq a+b\leq4,$$ which gives $$ab\leq4.$$
Id est, $$x^{\ln{y}}=e^{\ln{x}\ln{y}}=e^{ab}\leq e^4.$$
The equality occurs for $x=y=e^2,$ which says that we got a maximal value.
C-S it's the following.

For any $a_i$ and $b_i$ we have:
  $$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2.$$

In our case $n=2$, $a_1=a_2=1$, $b_1=a$ and $b_2=b$.
Thus, $$2(a^2+b^2)=(1^2+1^2)(a^2+b^2)\geq(1\cdot a+1\cdot b)^2=(a+b)^2.$$
By the way, also, you can get the last inequality by the following way:
$$2(a^2+b^2)-(a+b)^2=(a-b)^2\geq0.$$
</s>