<s>[INST] If $x,y>1$ and $(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2.$ then maximum value of $x^{\ln y}$ If $x,y>1$ and $(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2.$ then maximum value of $x^{\ln y}$
we can write $(\ln x)^2-\ln (x^2) = -((\ln y)^2-\ln (y^2))$ and assuming $x\geq y$
now let $f(x)=(\ln x)^2-\ln (x^2) $
$\displaystyle f'(x)=2 \ln x \cdot \frac{1}{x}-\frac{1}{x^2}\cdot 2x =\frac{2(\ln x-1)}{x}$
and $\displaystyle f''(x) = \frac{2(2-\ln x)}{x^2}$
wan,t be able to go further, could some help me with this, thanks
 [/INST] Let $\ln x = a$ and $\ln y=b$. Then the equation is written as
$$a^2+b^2 = 2a + 2b.$$
We can now find maximum values of a and b by making up a quadratic in $a$
and $b$, i.e.
$$
(a-1)^2 + (b-1)^2 = 2.
$$
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