Áp dụng bất đẳng thức Cô si cho hai số dương ta có:

(a2 + b2) + (b2 + c2) + (c2 + a2) ≥ 2ab + 2bc + 2ca

=> 2(a2 + b2 + c2 ) ≥ 2 (ab + bc + ca) (1) (a2 + 1) + (b2 + c2) + (c2 + a2) ≥ 2a + 2b + 2c

=> a2 + b2 + c2 + 3 ≥ 2(a + b + c) (2)

Cộng các vế của (1) và (2) ta có:

3 ( a2 + b2 + c2 ) + 3 ≥ 2 (ab + bc + ca + a + b + c)

=> 3( a2 + b2 + c2 ) + 3 ≥ 12 => a2 + b2 + c2 ≥ 3.

Ta có: (a^3/b + ab ) + ( b^3/c + bc ) + ( c^3/a + ca)≥ 2(a2 + b2 + c2) (CÔ SI) 

<=>a^3/b + b^3/c + c^3/a +ab + bc + ac  ≥ 2(a2 + b2 + c2)

Vì a2 + b2 + c2 ≥ ab + bc + ca => a^3 + b^3 + c^3 ≥ a2 + b2 + c2 ≥ 3 (đpcm).

Áp dụng bất đẳng thức cô-si cho hai số dương ta có:

\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2ab+2bc+2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\) (1)

\(\left(a^2+b^2\right)+\left(b^2+c^2\right)+\left(c^2+a^2\right)\ge2a+2b+2c\)

\(\Rightarrow a^2+b^2+c^2+3\ge2\left(a+b+c\right)\) (2)

Cộng (1) với (2)

\(3\left(a^2+b^2+c^2\right)+3\ge2\left(ab+bc+ca+a+b+c\right)\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)

\(\Rightarrow a^2+b^2+c^2\ge3\)

Ta có: \(\left(\dfrac{a^3}{b}+ab\right)+\left(\dfrac{b^3}{c}+bc\right)+\left(\dfrac{c^3}{a}+ca\right)\ge2\left(a^2+b^2+c^2\right)\)

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

Vì \(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge a^2+b^2+c^2\ge3\) (đpcm).

\(ab+bc+ac=3\)

Ta có:

\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\ge\dfrac{2}{ab+1}\) ( đúng với mọi \(ab\ge1\))

Giả sử:\(ab\ge1\)

\(\Rightarrow\dfrac{2}{ab+1}+\dfrac{1}{c^2+1}\ge\dfrac{2c^2+2+ab+1}{\left(ab+1\right)\left(c^2+1\right)}=\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\)

Giả sử: \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\)(đúng)

\(\Leftrightarrow2\left(2c^2+ab+3\right)\ge3\left(ab+1\right)\left(c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3\left(abc^2+ab+c^2+1\right)\)

\(\Leftrightarrow4c^2+2ab+6\ge3abc^2+3ab+3c^2+3\)

\(\Leftrightarrow c^2-ab-3abc^2+3\ge0\)

\(\Leftrightarrow c^2-ab-3abc^2+ab+ac+bc\ge0\)   ( vì \(ab+ac+bc=3\) )

\(\Leftrightarrow c^2+ac+bc-3abc^2\ge0\)

\(\Leftrightarrow c+a+b-3abc\ge0\)

\(\Leftrightarrow c+a+b\ge3abc\)

Ta có:

\(3\left(c+a+b\right)=\left(ab+ac+bc\right)\left(c+a+b\right)\) ( vì \(ab+ac+bc=3\) )

Áp dụng BĐT AM-GM, ta có:

\(\left(ab+ac+bc\right)\left(c+a+b\right)\ge9abc\)

\(\Rightarrow a+b+c\ge3abc\)

\(\Rightarrow\) \(\dfrac{2c^2+ab+3}{\left(ab+1\right)\left(c^2+1\right)}\ge\dfrac{3}{2}\) ( luôn đúng )

\(\Rightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}\) ( đfcm )

Dấu "=" xảy ra khi \(a=b=c=1\)

Hình như sai đề rồi bạn ạ, dấu ≥ phải là ≤

Lời giải:

\(\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c\)

\(\Leftrightarrow \frac{a^2+bc}{b+c}-c+\frac{b^2+ac}{a+c}-a+\frac{c^2+ab}{a+b}-b\geq 0\)

\(\Leftrightarrow \frac{a^2-c^2}{b+c}+\frac{b^2-a^2}{a+c}+\frac{c^2-b^2}{a+b}\geq 0\)

\(\Leftrightarrow a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\geq 0\)

\(\Leftrightarrow \frac{a^2(a-b)(a+b)+b^2(b-c)(b+c)+c^2(c-a)(c+a)}{(a+b)(b+c)(c+a)}\geq 0\)

\(\Leftrightarrow a^2(a^2-b^2)+b^2(b^2-c^2)+c^2(c^2-a^2)\geq 0\)

\(\Leftrightarrow a^4+b^4+c^4-(a^2b^2+b^2c^2+c^2a^2)\geq 0\)

\(\Leftrightarrow \frac{(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2}{2}\geq 0\) (luôn đúng)

Do đó ta có đpcm

Dấu bằng xảy ra khi $a=b=c$

\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)

\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)

\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ 

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

\(\Rightarrow VT=\dfrac{1}{\dfrac{y}{x}\left(\dfrac{z}{y}+1\right)}+\dfrac{1}{\dfrac{z}{y}\left(\dfrac{x}{z}+1\right)}+\dfrac{1}{\dfrac{x}{z}\left(\dfrac{y}{x}+1\right)}\)

\(VT=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)

\(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\dfrac{3}{2}\)

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

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

\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)

\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{b}{b^2+a}+\dfrac{c}{c^2+b}+\dfrac{a}{a^2+c}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{b}{b^2+a}\le\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)

Thiết lập tương tự và thu lại tao có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\sqrt{\dfrac{1}{a}}\le\dfrac{\dfrac{1}{a}+1}{2}\)

Tương tự ta có

\(\sqrt{\dfrac{1}{b}}\le\dfrac{\dfrac{1}{b}+1}{2};\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{c}+1}{2}\)

Thu lại ta có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\right)\)

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}-\dfrac{3}{4}=\dfrac{3}{2}\)

\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=1\)