Đặt $I=\int\limits_0^{\pi/2}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx,J=\int\limits_0^{\pi/2}\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$
Đặt: $x=\dfrac{\pi}{2}-t\Rightarrow dx=-dt$
Ta có:
$I=-\int\limits_{\pi/2}^0\dfrac{\sqrt{\sin(\displaystyle\dfrac{\pi}{2}-t)}}{\sqrt{\sin(\displaystyle\dfrac{\pi}{2}-t)}+\sqrt{\cos(\displaystyle\dfrac{\pi}{2}-t)}}dt$
   $=\int\limits_0^{\pi/2}\dfrac{\sqrt{\cos t}}{\sqrt{\sin t}+\sqrt{\cos t}}dt=J$
Mà ta có:
$I+J=\int\limits_0^{\pi/2}dx=\dfrac{\pi}{2}$
Suy ra: $I=J=\dfrac{\pi}{4}$

b. Bất phương trình tương đương với:
       $(x+1)(x+4)<5\sqrt{x^2+5x+28}$
$\Leftrightarrow \left[\begin{array}{l}(x+1)(x+4)\le0\\\left\{\begin{array}{l}(x+1)(x+4)>0\\(x+1)^2(x+4)^2<25(x^2+5x+28)\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}-4\le x\le-1\\\left\{\begin{array}{l}(x+1)(x+4)>0\\x^4+10x^3+8x^2-85x-684<0\end{array}\right.\end{array}\right.$
 $\Leftrightarrow -9<x<4$

a.
Điều kiện: $x(x+3)\ge0 \Leftrightarrow \left[\begin{array}{l}x\ge0\\x\le-3\end{array}\right.$
Bất phương trình tương đương với:
       $3\sqrt{x(x+3)}>(x+5)(2-x)$
$\Leftrightarrow \left[\begin{array}{l}(x+5)(2-x)\le0\\\left\{ \begin{array}{l}(x+5)(2-x)>0\\9x(x+3)>(x+5)^2(2-x)^2\end{array} \right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge2\\x\le-5\\\left\{\begin{array}{l}(x+5)(2-x)>0\\x^4+6x^3-20x^2-87x+100<0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge2\\x\le-5\\\left\{\begin{array}{l}(x+5)(2-x)>0\\(x-1)(x+4)(x^2+3x-25)<0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x<-4\\x>1\end{array}\right.$

b. Điều kiện: $2x^2-3x-5\ge0 \Leftrightarrow \left[\begin{array}{l}x\ge\dfrac{5}{2}\\x\le-1\end{array}\right.$
Bất phương trình tương đương với:
     $\sqrt{2x^2-3x-5}< x-1$
$\Leftrightarrow \left\{\begin{array}{l}x-1>0\\2x^2-3x-5\ge0\\2x^2-3x-5<(x-1)^2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x\ge\dfrac{5}{2}\\x^2-x-6<0\end{array}\right.\Leftrightarrow \dfrac{5}{2}\le x<3$

a.
Điều kiện: $5\le x\le 9$
Bất phương trình tương đương với:
     $\sqrt{x-5}>\sqrt{9-x}+1$
$\Leftrightarrow x-5>9-x+2\sqrt{9-x}+1$
$\Leftrightarrow 2x-15>2\sqrt{9-x}$
$\Leftrightarrow \left\{\begin{array}{l}2x-15>0\\9-x\ge0\\(2x-15)^2>4(9-x)\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}\dfrac{15}{2}<x\le9\\4x^2-56x+189>0\end{array}\right.\Leftrightarrow \dfrac{1}{2}(14+\sqrt7)<x\le9$

b. Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow \begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4 \end{matrix}} \right. \end{cases}\Leftrightarrow x>4$
Bất phương trình tương đương với:
     $\dfrac{\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}  > \dfrac{5}{\sqrt{x-3}}$
$\Leftrightarrow \sqrt{x^{2}-16}+x-3  >5$
$\Leftrightarrow \sqrt{x^{2}-16} > 8-x$
$\Leftrightarrow \left[\begin{array}{l}8-x\le0\\\begin{cases}8-x>0\\x^{2}-16>(8-x)^2\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge8\\\begin{cases}x<8\\x^2-16>x^2-16x+64\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge8\\\begin{cases}x<8\\x>5\end{cases}\end{array}\right. \Leftrightarrow x>5$

a. Điều kiện: $x\le 2$
Bất phương trình tương đương với:
     $\dfrac{4}{\sqrt{2-x}}-\dfrac{2-x}{\sqrt{2-x}}<2$
$\Leftrightarrow x+2<2\sqrt{2-x}$
$\Leftrightarrow \left[\begin{array}{l}x+2\le0\\\left\{\begin{array}{l}x+2>0\\(x+2)^2<4(2-x)\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\le-2\\\left\{\begin{array}{l}x>-2\\x^2+8x-4<0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\le-2\\\left\{\begin{array}{l}x>-2\\-4-2\sqrt5<x<-4+2\sqrt5\end{array}\right.\end{array}\right.\Leftrightarrow x<-4+2\sqrt5$

Điều kiện $\begin{cases}x>3\\ x^2 \ge 16 \end{cases}\Leftrightarrow \begin{cases}x>3 \\ \left[ {\begin{matrix} x\ge4\\ x\le-4 \end{matrix}} \right. \end{cases}\Leftrightarrow x>4$
BPT
$\Leftrightarrow \dfrac{2\sqrt{x^{2}-16}}{\sqrt{x-3}}+\dfrac{x-3}{\sqrt{x-3}}  > \dfrac{7-x}{\sqrt{x-3}}$
$\Leftrightarrow 2\sqrt{x^{2}-16}+x-3  > 7-x$
$\Leftrightarrow 2\sqrt{x^{2}-16} > 10-2x$
$\Leftrightarrow \sqrt{x^{2}-16} > 5-x$
$\Leftrightarrow \left[\begin{array}{l}5-x\le0\\\begin{cases}5-x>0\\x^{2}-16>(5-x)^2\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x^2-16>x^2-10x+25\end{cases}\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x\ge5\\\begin{cases}x<5\\x>\dfrac{41}{10}\end{cases}\end{array}\right. \Leftrightarrow x>\dfrac{41}{10}$

Ta có:
$A=\dfrac{abc}{a^3(b+c)}+\dfrac{abc}{b^3(c+a)}+\dfrac{abc}{c^3(a+b)}$
    $=\dfrac{\dfrac{1}{a^2}}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{\dfrac{1}{b^2}}{\dfrac{1}{c}+\dfrac{1}{a}}+\dfrac{\dfrac{1}{c^2}}{\dfrac{1}{a}+\dfrac{1}{b}}$
    $\ge\dfrac{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}{2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}$
    $=\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$
    $\ge\dfrac{3}{2}\sqrt[3]{\dfrac{1}{abc}}=\dfrac{3}{2}$
Dấu bằng xảy ra khi: $a=b=c=1$

Bạn xem lời giải ở phần (b) bài này nhé.

http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/115185/tich-phan-dac-biet-10

Ta có :
    $\cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2} =\dfrac{1}{2}\cos^2\dfrac{A}{2}\left[\cos\dfrac{B+C}{2}+\cos\dfrac{B-C}{2}\right]$
Vì thế :
    $\cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\le\dfrac{1}{2}\cos^2\dfrac{A}{2}(1+\sin\dfrac{A}{2})$
$\Leftrightarrow \cos^2\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\le\dfrac{1}{4}(1+\sin\dfrac{A}{2})(1+\sin\dfrac{A}{2})(2-2\sin\dfrac{A}{2})$
Áp dụng bất đẳng thức Cauchy, ta có :
  $(1+\sin\dfrac{A}{2})(1+\sin\dfrac{A}{2})(2-2\sin\dfrac{A}{2})\le\left[\dfrac{(1+\sin\dfrac{A}{2})+(1+\sin\dfrac{A}{2}) + (2-2\sin\dfrac{A}{2})}{3} \right]^3$
$ \Rightarrow (1+\sin\dfrac{A}{2})^2(2-2\sin\dfrac{A}{2}) \le \dfrac{64}{27}$
Suy ra:
$\cos\dfrac{A}{2}\sqrt{\cos\dfrac{B}{2}\cos\dfrac{C}{2}} \le \dfrac{4\sqrt3}{9}$
Dấu bằng xảy ra khi  $B=C,A = 2\arcsin\dfrac{1}{3}$