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 ?

From the given conditions we can write, $a=\sin \theta, b=\cos \theta,\ \theta\in (0,\pi/2)$. Then, the objective function becomes $$f(\theta)=25/2\sin 2\theta+25\sin \theta+10\cos\theta\\ f'(\theta)=25\cos 2\theta+25\cos\theta-10\sin \theta$$Equate this to $0$ to get a solution of $\theta$, I believe it pertains to solving a cubic equation $100\cos^3\theta-71\cos\theta+21=0$ which results in three real solutions one of which, that maximizes the function, is $\cos \theta=\frac{3}{5},\implies \sin \theta=\frac{4}{5},\ \sin 2\theta=\frac{24}{25}\implies f(\theta)\le 38,\ \forall \theta\in (0,\pi/2)$.
