Prove this inequality $25ab+25a+10b\le38$ let $a,b>0$,and such $a^2+b^2=1$,show that
$$25ab+25a+10b\le38$$
Now I have found this inequality $"="$,if and only if $a=\dfrac{4}{5},b=\dfrac{3}{5}$
then  How to prove this inequality by AM-GM or other ?

By setting $a=\cos\theta,b=\sin\theta$ you just have to prove that for any $\theta\in\left(0,\frac{\pi}{2}\right)$ we have:
$$ f(\theta)=(5a+2)(5b+5)=(5\cos\theta+2)(5\sin\theta+5)\leq 48,$$
or, by using Weierstrass substitution $\theta=2\arctan t$,
$$\forall t\in(0,1),\quad g(t) = (7-3t^2)(t+1)^2- \frac{48}{5}(1+t^2)^2 \leq 0$$
that is straightforward to check since:
$$ g'(t) = \frac{2}{5}(1-3t)\left(35+29 t+42 t^2\right), $$
hence we have equality only for $t=\frac{1}{3}$, i.e. for $\theta=2\arctan\frac{1}{3}=\arctan\frac{3}{4}=\arcsin\frac{3}{5}$.
