Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

Áp dụng BĐT cosi:

\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

\(=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}\)

\(=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}\)

\(=\frac{a^2-2a+1}{\left(1-a\right)^2}=\frac{\left(a-1\right)^2}{\left(1-a\right)^2}\)

\(=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(ĐPCM\right)\)

Lời giải:

Bổ sung điều kiện $a,b$ là các số dương. Áp dụng BĐT Cô-si ta có:

$a+b\geq 2\sqrt{ab}$

$\frac{1}{a}+\frac{1}{b}\geq 2\sqrt{\frac{1}{ab}}$

$\Rightarrow (a+b)(\frac{1}{a}+\frac{1}{b})\geq 2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4$

Ta có đpcm 

Dấu "=" xảy ra khi $a=b$

\(\Leftrightarrow2a^3+2b^3-a^3-ab^2-a^2b-b^3>=0\)

\(\Leftrightarrow a^3+b^3-ab^2-a^2b>=0\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)>=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2>=0\)(luôn đúng)

Ta có : 

          \(\log_ab\ge\log_{a+c}\left(b+c\right)\Leftrightarrow\log_ab-1\ge\log_{a+c}\left(b+c\right)-1\)

                                          \(\Leftrightarrow\log_a\frac{b}{a}\ge\log_{a+c}\frac{b+c}{a+c}\)  

Với \(1< a\le b\) và \(c\ge0\Rightarrow\frac{b}{a}\ge\frac{b+c}{a+c}\ge1\) nên \(\log_a\frac{b}{a}\ge\log_a\frac{b+c}{a+c}\) (*)

Mặt khác, ta được : \(\log_a\frac{b+c}{a+c}\ge\log_{a+c}\frac{b+c}{a+c}\)  (**)

Từ (*) và (**) \(\Rightarrow\log_ab\ge\log_{a+c}\left(b+c\right)\)

Dấu "=" xảy ra khi c = 0 hoặc a = b

1)))))))

\(\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{\left(\sqrt{ab}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2}{\sqrt{ab}}.\frac{\left(\sqrt{ab}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{2\sqrt{ab}-a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

\(=\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1\)

\(\text{VT}=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)=\left(1+\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)

\(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=\text{VP(điều phải chứng minh)}\)