Áp dụng BĐT AG-GM:

\(\dfrac{a^3}{a^2+ab+b^2}\ge\dfrac{a^3}{a^2+\dfrac{a^2+b^2}{2}+b^2}=\dfrac{a^3}{\dfrac{3}{2}\left(a^2+b^2\right)}\)

Cmtt \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+bc+c^2}\ge\dfrac{b^3}{\dfrac{3}{2}\left(b^2+c^2\right)}\\\dfrac{c^3}{c^2+ac+a^2}\ge\dfrac{c^3}{\dfrac{3}{2}\left(c^2+a^2\right)}\end{matrix}\right.\)

Cộng vế theo vế của bất đẳng thức:

\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\)

Tiếp tục áp dụng BĐT AG-GM:

\(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)

Cmtt\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^3}{b^2+c^2}\ge b-\dfrac{c}{2}\\\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\end{matrix}\right.\)

Cộng vế theo vế

\(\Leftrightarrow VT\ge\dfrac{2}{3}\left(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\right)\\ \ge\dfrac{2}{3}\left(a-\dfrac{b}{2}+b-\dfrac{c}{2}+c-\dfrac{a}{2}\right)=\dfrac{2}{3}\left(a+b+c-\dfrac{a+b+c}{2}\right)=\dfrac{a+b+c}{3}\)

\(\dfrac{a^3}{a^2+ab+b^2}=a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\dfrac{ab\left(a+b\right)}{3\sqrt[3]{a^2.ab.b^2}}=a-\dfrac{a+b}{3}=\dfrac{2a-b}{3}\)

Tương tự và cộng lại ta sẽ có đpcm

a)Bunhia:

\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)

b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bđt câu a

=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)

\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)

Tự tìm dấu "="

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c=c.1 thay 1 bằng a+b+c xong cô si

Áp dụng bđt cosi schwart ta có:

`VT>=(a+b+c)^2/(a+b+c+sqrt{ab}+sqrt{bc}+sqrt{ca})`

Dễ thấy `sqrt{ab}+sqrt{bc}+sqrt{ca}<a+b+c`

`=>VT>=(a+b+c)^2/(2(a+b+c))=(a+b+c)/2=3`

Dấu "=" `<=>a=b=c=1.`

Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).

Đặt biểu thức ở VT là A. Ta có: 

\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).

Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).

Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.

Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).

Đẳng thức xảy ra khi a = b = c = 3.

Vậy...

Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)

\(VT\ge\dfrac{\sqrt{\dfrac{1}{3}\left(a+a+b\right)^2}}{ab}+\dfrac{\sqrt{\dfrac{1}{3}\left(b+b+c\right)^2}}{bc}+\dfrac{\sqrt{\dfrac{1}{3}\left(c+c+a\right)^2}}{ca}\)

\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Áp đụng bất đẳng thức AM - GM, ta có:

\(\dfrac{a^3}{b}+ab\ge2a^2\)

\(\dfrac{b^3}{c}+bc\ge2b^2\)

\(\dfrac{c^3}{a}+ca\ge2c^2\)

Theo hệ quả của bất đẳng thức AM - GM thì:

\(a^2+b^2+c^2\ge ab+bc+ac\)

Do đó, \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\left(đpcm\right)\)

Dấu "=" xảy ra khi a = b = c = 0.

\(\dfrac{P}{\sqrt{2}}=\dfrac{a}{\sqrt{2b\left(a+b\right)}}+\dfrac{b}{\sqrt{2c\left(b+c\right)}}+\dfrac{c}{\sqrt{2a\left(a+c\right)}}\)

\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2a}{2b+a+b}+\dfrac{2b}{2c+b+c}+\dfrac{2c}{2a+a+c}\)

\(\dfrac{P}{\sqrt{2}}\ge2\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)=2\left(\dfrac{a^2}{a^2+3ab}+\dfrac{b^2}{b^2+3bc}+\dfrac{c^2}{c^2+3ca}\right)\)

\(\dfrac{P}{\sqrt{2}}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+ab+bc+ca}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\dfrac{1}{3}\left(a+b+c\right)^2}=\dfrac{3}{2}\)

\(\Rightarrow P\ge\dfrac{3\sqrt{2}}{2}\) (đpcm)

\(\dfrac{a}{\sqrt{ab+b^2}}=\dfrac{\sqrt{2}.a}{\sqrt{2b\left(a+b\right)}}\ge\dfrac{\sqrt{2}.a}{\dfrac{2b+a+b}{2}}=\dfrac{2\sqrt{2}a}{a+3b}\)

làm tương tự với \(\dfrac{b}{\sqrt{bc+c^2}};\dfrac{c}{\sqrt{ca+a^2}}\)

\(=>P\ge2\sqrt{2}\left(\dfrac{a}{a+3b}+\dfrac{b}{b+3c}+\dfrac{c}{c+3a}\right)\)

\(=2\sqrt{2}\left(\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ca\right)}\right)\)

\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\dfrac{4}{3}\left(ab+bc+ca\right)+\dfrac{8}{3}\left(ab+bc+ca\right)}\right]\)

\(=2\sqrt{2}\left[\dfrac{\left(a+b+c\right)^2}{\dfrac{4}{3}\left(a+b+c\right)^2}\right]=\dfrac{2\sqrt{2}.3}{4}=\dfrac{3\sqrt{2}}{2}\)

dấu"=" xảy ra<=>a=b=c