Show that if $ 0\leq x\leq\pi/2 $ $$2^{\sqrt{\sin x}}+2^{\sqrt{\cos x}}\geq 2+\sin x+\cos x \geq 3$$ I encountered this problem when I tried to solve the equation $$2^{\sqrt{\sin x}}+2^{\sqrt{\cos x}}=2+\sin x+\cos x$$ I tried to use $A.M\geq G.M$ but it didn't help or I could not solve it properly.
Inequality involving trigonometry and exponentiation
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trigonometry
inequality
exponential-function
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1Set $x=\frac\pi4$ to check the last inequality – 2017-02-06
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0In the range you have $0\le \cos x \le 1$ so that $\cos x \le \sqrt {\cos x}$. Have you tried the binomial theorem with $2^{\sqrt {\cos x}}=(1+1)^{\sqrt {\cos x}}$? – 2017-02-06
1 Answers
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Hint: for $0\leqslant x\leqslant1$ $$2^x\geqslant 1 + x^2$$
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0And for the second part, $$(\sin x+\cos x)^2 = 1+\sin(2x) \geq 1.$$ – 2017-02-06
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0@JackD'Aurizio Very good, Thanks. – 2017-02-06
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0I forgot a (+1) before. It may be interesting to point out that such inequality follows from the concavity of the function $2^{\sqrt{x}}$ over $(0,1)$. – 2017-02-06