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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:

$$a+b\geq 2\sqrt{1-t^2}\sqrt{ab}+(a-b)t$$

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    I suppose $t \in [0,1]$? – 2017-02-17
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    a known proof for the generalized form uses induction – 2017-02-17
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    Of course!!!!!!!!!!!!!!!!!!!!!!!!!!! – 2017-02-17
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    Did you use calculus or I have a way – 2017-02-17
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    I didn't use calculus i used Cauchy-Schwarz inequality and trig functions – 2017-02-17

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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.

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Your inequality is wrong! Try $t=2$.

For $-1\leq t\leq 1$ we need to prove that $$(1-t)a+(1+t)b\geq2\sqrt{(1-t)(1+t)ab},$$ which is just AM-GM

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    I do not even need to say that $-1 \leq t \leq 1$,this is obviously, the inequality isn't wrong, Nobody here said that t can be any real... – 2017-02-17
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    @Israel Meireles Chrisostomo You said that "Let a,b positive real numbers, and t the parameter.", which says that $a>0$, $b>0$ and $t$ is a real number. I think it would better if you'll read my proof for $-1\leq t\leq1$. – 2017-02-17