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 ?

Given the solution $a=4/5,b=3/5$ gives equality, try $$a=\frac45\cos\theta-\frac35\sin\theta,b=\frac35\cos\theta+\frac45\sin\theta$$
(sorry, my comment was wrong at first) I think it becomes
$$12(\cos^2\theta-\sin^2\theta)+26\cos\theta-7\sin\theta+7\cos\theta\sin\theta)\leq38\\
24(\cos^2\theta-1)+26(\cos\theta-1)+7\sin\theta(\cos\theta-1)\leq0\\
(\cos\theta-1)(50+24\cos\theta+7\sin\theta)\leq0 $$
which of course is true, equality if $\theta=0$
