$a\sqrt{\sin x}+b\sqrt{\cos x}≤(a^{4/3}+b^{4/3})^{3/4}$

Question

prove that : $$a,b>0\\,0

$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\frac{a}{b}\sqrt{\cos x})$$

$$\frac{a}{b}=\tan y$$

$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\tan y\sqrt{\cos x})$$

$$\frac{\sin y}{\cos y}=\tan y$$

$$a\sqrt{\sin x}+b\sqrt{\cos x}=a(\sqrt{\sin x}+\frac{\sin y}{\cos y}\sqrt{\cos x})$$

now?

Answer 1

You should apply Hölder inequality, which asserts that given positive $2n$ real numbers $a_1,\cdots,a_n$, $b_1,\cdots,b_n$ and "conjugate exponents" $p,q$ (i.e. positive real numbers such that $\frac 1p+\frac 1q=1$) :

$$\sum_{i=1}^na_ib_i\le\left(\sum_{i=1}^na_i^p\right)^{1/p}\left(\sum_{i=1}^nb_i^q\right)^{1/q}$$

Here we choose $n=2$, $(a_1,a_2)=(a,b)$, $(b_1,b_2) =(\sqrt{\cos(x)},\sqrt{\sin(x)})$, and $(p,q)=(\frac43,4)$ and get :

$$a\sqrt{\cos(x)}+b\sqrt{\sin(x)}\le\left(a^{4/3}+b^{4/3}\right)^{3/4}(\cos^2(x)+\sin^2(x))^{1/4}=\left(a^{4/3}+b^{4/3}\right)^{3/4}$$

Hence the conclusion.

Answer 2

Using AM-GM inequality,

$a^{4/3}+b^{4/3} \ge 2\sqrt{a^{4/3}b^{4/3}}$

$a^{4/3}+b^{4/3} \ge 2ab^{2/3}$

$\left(a^{4/3}+b^{4/3}\right)^{3/4} \ge 2^{3/4}{ab^{1/2}} \cdots(1)$

Using AM-GM inequality,

$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2\sqrt{ab\sqrt{\sin x \cos x}}$

$a\sqrt{\sin x}+b\sqrt{\cos x} \ge 2(ab)^{1/2}(\sin x\cos x)^{1/4} \cdots(2)$

Divide the equations

$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge 2^{1/4}(\sin x\cos x)^{1/4}$

$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge 2^{1/4}\left[\dfrac{\sin(2x)}{2}\right]^{1/4}$

$\dfrac{a\sqrt{\sin x}+b\sqrt{\cos x}}{\left(a^{4/3}+b^{4/3}\right)^{3/4}} \ge [\sin(2x)]^{1/4} \cdots(3)$

Note that $0 \le \sin(2x) \le 1$

or

$0 \le [\sin(2x)]^{1/4} \le 1$

Therefore, numerator of equation (3) must be less than or equal to denominator

i.e., $a\sqrt{\sin x}+b\sqrt{\cos x} \le \left(a^{4/3}+b^{4/3}\right)^{3/4}$