Inequality of arithmetic and geometric mean I did a proof for inequality below, anyone has a other proof?
Let $a$ and $b$ be positive real numbers, and $t$ the parameter. Prove that:
$$a+b\geq 2\sqrt{1-t^2}\sqrt{ab}+(a-b)t$$

Let $f (t)=Rhs $ of inequality . So differentiating and setting it equal to $0$ gives $\frac {2t}{\sqrt {1-t^2}}=\frac {a-b}{\sqrt {ab}} $ squaring both sides and solving we have $t=\frac {a-b}{a+b} $ thus putting the value of $t $ in original equation and simplifying we get it as $a^2+b^2$ now both $a,b $ are positive so $a^2+b^2\leq (a+b)^2$. Thus its proved. $$\text {another way} $$ put $t=\sin (x) $ thus $\sqrt {1-t^2}=\cos (x) $ we can do this as for the equality to give real values t has to be in $[0,1] $ thus we have $f (x)=2\sqrt {ab}\cos (x)+(a-b)\sin (x)\leq \sqrt {(2\sqrt {ab})^2+(a-b)^2}$  which is same as $\sqrt {(a+b)^2}=(a+b) $ as both $a,b $ are positive.
