Ta có: \(\left(a+b\right)^2\ge4ab\)

Từ đó ta có

\(\left(a+b\right)^2+\frac{a+b}{2}\ge4ab+\frac{a+b}{2}\)

Ta cần chứng minh 

\(4ab+\frac{a+b}{2}\ge2a\sqrt{b}+2b\sqrt{a}\)

\(\Leftrightarrow8ab+a+b-4a\sqrt{b}-4b\sqrt{a}\ge0\)

\(\Leftrightarrow\left(4ab-4a\sqrt{b}+a\right)+\left(4ab-4b\sqrt{a}+b\right)\ge0\)

\(\Leftrightarrow\left(2\sqrt{ab}-\sqrt{a}\right)^2+\left(2\sqrt{ab}-\sqrt{b}\right)^2\ge0\)(đúng)

\(\Rightarrow\)ĐPCM là đúng

Bunhiacopxki:

\(\left(b+a+a\right)\left(b+c+\dfrac{c^2}{a}\right)\ge\left(b+\sqrt{ca}+c\right)^2\)

\(\Rightarrow\dfrac{2a^2+ab}{\left(b+\sqrt{ca}+c\right)^2}\ge\dfrac{2a^2+ab}{\left(2a+b\right)\left(b+c+\dfrac{c^2}{a}\right)}=\dfrac{a^2}{c^2+ab+bc}\)

Tương tự:

\(\dfrac{2b^2+bc}{\left(c+\sqrt{ca}+a\right)^2}\ge\dfrac{b^2}{a^2+ab+bc}\)

\(\dfrac{2c^2+ca}{\left(a+\sqrt{bc}+b\right)^2}\ge\dfrac{c^2}{b^2+ac+bc}\)

\(\Rightarrow P\ge\dfrac{a^2}{c^2+ab+ac}+\dfrac{b^2}{a^2+ab+bc}+\dfrac{c^2}{b^2+ac+bc}\)

\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

Dấu "=" xảy ra khi \(a=b=c\)

Câu b : Ta có :

\(\left(a+b\right)^2+\dfrac{a+b}{2}=\left(a+b\right)\left(a+b+\dfrac{1}{2}\right)=\left(a+b\right)\left[\left(a+\dfrac{1}{4}\right)+\left(b+\dfrac{1}{4}\right)\right]\)

Áp dụng BĐT Cô - Si ta có :

\(\left\{{}\begin{matrix}a+b\ge2\sqrt{ab}\\a+\dfrac{1}{4}\ge\sqrt{a}\\b+\dfrac{1}{4}\ge\sqrt{b}\end{matrix}\right.\)

\(\Rightarrow VT\ge2\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)=2a\sqrt{b}+2b\sqrt{a}\) ( đpcm )

Dấu \("="\) xảy ra khi \(a=b=-\dfrac{1}{4}\)

Ta có 

\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{b}+\frac{1}{a}\right)\)

                                                                \(\ge3+2a.\frac{1}{\sqrt{bc}}+2b.\frac{1}{\sqrt{ac}}+2c.\frac{1}{\sqrt{ab}}\)

Mà \(abc\le1\)

=> \(VT\ge3+2a\sqrt{a}+2b\sqrt{b}+2c\sqrt{c}=VP\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

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