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 ?
 A: I think we can use AM-GM and Cauchy-Schwarz inequality to solve this problem
Note
$$25ab+25a+10b=(5a-4)(5b-3)+40a+30b-12$$
Use AM-GM inequality we have
$$(5a-4)(5b-3)\le \dfrac{1}{2}[(5a-4)^2+(5b-3)^2]=\dfrac{1}{2}[50-40a-30b]$$
so
$$25ab+25a+10b\le 20a+15b+13=5[4a+3b]+13$$
and Use Cauchy-Schwarz inequality we have
$$25=[4^2+3^2][a^2+b^2]\ge (4a+3b)^2$$
so we have
$$25ab+25a+10b\le 5\cdot 5+13=38$$
A: Just another way - once you know the equality conditions, all you need to make sure is to use inequalities which maintain them.  In this case, you need to find bounds for $ab, a, b$ in terms of $a^2, b^2$.  So the following inequalities look promising
$$(3a-4b)^2 \ge 0, \quad (5a-4)^2 \ge 0, \quad (5b-3)^2 \ge 0$$
These give the bounds:
$$24ab \le 9a^2+16b^2, \quad 40a \le 16+25a^2, \quad 30b \le 9+25b^2$$
Putting it together, we have
$$25ab+25a+10b \le \frac{25}{24}(9a^2+16b^2)+\frac{25}{40}(16+25a^2)+\frac{10}{30}(9+25b^2) \\= 13+25(a^2+b^2)=38 $$
A: 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 $\cos \theta=\frac{3}{5},\implies \sin \theta=\frac{4}{5},\ \sin 2\theta=\frac{24}{25}\implies f(\theta)\le 38,\ \forall \theta\in (0,\pi/2)$.
A: Let's write $a=(4+u)/5$ and $b=(3-v)/5$.  The condition $a^2+b^2=1$ becomes
$$u^2+8u-6v+v^2=0$$
The inequality $25ab+25a+10b\le38$ becomes
$$8u-6v\le uv$$
Putting these together, the inequality is equivalent to
$$0\le u^2+uv+v^2$$
which holds for all $u$ and $v$.
Note, it's easy to see from $a^2+b^2=1$ that $u$ and $v$ cannot have opposite signs, which makes the equivalent inequality completely obvious, but the inequality holds in general.
A: 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 $\theta=2\arctan\frac{1}{3}=\arctan\frac{3}{4}=\arcsin\frac{3}{5}$.
A: 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 $\theta=0$
