\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\Leftrightarrow\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}\le1\)

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

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

\(\Rightarrow\) đpcm

\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{c+a+1}\ge1\)

\(\Leftrightarrow2\ge\dfrac{a+b}{a+b+1}+\dfrac{b+c}{b+c+1}+\dfrac{c+a}{c+a+1}=\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+a+b}+\dfrac{\left(b+c\right)^2}{\left(b+c\right)^2+b+c}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)^2+c+a}\)

\(\Rightarrow2\ge\dfrac{2\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca+a+b+c}\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)+2\left(a+b+c\right)\ge2\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)\)

\(\Rightarrow\)đpcm

\(VT\ge\dfrac{1}{\left(a^2+1\right)-1}+\dfrac{1}{\left(b^2+1\right)-1}+\dfrac{1}{\left(c^2+1\right)-1}+4-\dfrac{4}{ab+1}+4-\dfrac{4}{bc+1}+4-\dfrac{4}{ca+1}\)

\(VT\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{4}{ab+1}-\dfrac{4}{bc+1}-\dfrac{4}{ca+1}+12\)

Mặt khác \(a;b;c\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab+1\ge a+b\) (và tương tự...)

\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+12\)

\(VT\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}-\dfrac{4}{a+b}-\dfrac{4}{b+c}-\dfrac{4}{c+a}+1+1+1+9\)

\(VT\ge\left(\dfrac{2}{a+b}-1\right)^2+\left(\dfrac{2}{b+c}-1\right)^2+\left(\dfrac{2}{c+a}-1\right)^2+9\ge9\)

\(ab+bc+ac=3\)

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 ≤

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}=\dfrac{a^2+b^2+2}{a^2b^2+a^2+b^2+1}=1-\dfrac{a^2b^2-1}{a^2b^2+a^2+b^2+1}\ge1-\dfrac{a^2b^2-1}{a^2b^2+2ab+1}\)

\(=1-\dfrac{\left(ab-1\right)\left(ab+1\right)}{\left(ab+1\right)^2}=1-\dfrac{ab-1}{ab+1}=\dfrac{2}{ab+1}\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

\(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\)

\(\Rightarrow\left(\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}\right)+\left(\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\right)\ge0\)

\(\Rightarrow\dfrac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{ab-b^2}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\dfrac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{a\left(b-a\right)\left(1+b^2\right)+b\left(a-b\right)\left(1+a^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)\left(a+ab^2\right)-\left(b-a\right)\left(b+a^2b\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)\left(-\left(b-a\right)+ab\left(b-a\right)\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Rightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng vì \(ab\ge1\))

\(VT=3\left(\dfrac{1}{4ab}+\dfrac{1}{a^2+4b^2}\right)+\dfrac{1}{2.a.2b}\ge\dfrac{12}{a^2+4ab+4b^2}+\dfrac{2}{\left(a+2b\right)^2}=14\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)

Bài toán cơ bản:

\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\) 

Bunhiacopxki:

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

\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Cách 2:

Do \(abc=1\), đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

Ta có \(\dfrac{a}{\left(ab+a+1\right)^2}=\dfrac{\dfrac{x}{y}}{\left(\dfrac{x}{z}+\dfrac{x}{y}+1\right)^2}=\dfrac{\dfrac{x}{y}.y^2z^2}{\left(xy+yz+zx\right)^2}=\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}\)...

Từ đó, BĐT cần chứng minh trở thành:

\(\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}+\dfrac{x^2yz}{\left(xy+yz+zx\right)^2}+\dfrac{xy^2z}{\left(xy+yz+zx\right)^2}\ge\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}}\)

\(\Leftrightarrow xyz\left(x+y+z\right)\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\ge\left(xy+yz+zx\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2z+y^2x+z^2y\right)\ge\left(xy+yz+zx\right)^2\)

Thật vậy, áp dụng BĐT Bunhiacopxki:

\(\left(z+x+y\right)\left(x^2z+y^2x+z^2y\right)\ge\left(\sqrt{zx^2z}+\sqrt{xy^2x}+\sqrt{yz^2y}\right)^2=\left(xy+yz+zx\right)^2\) (đpcm)

* Vì \(a,b\ge1\)nên \(\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab+1\ge a+b\)

Một cách tương tự: \(bc+1\ge b+c;ca+1\ge c+a\)

Với mọi số thực \(a\ge1\) ta luôn có: \(\left(a-1\right)^2\ge0\Leftrightarrow a^2\ge2a-1\Leftrightarrow\frac{1}{2a-1}\ge\frac{1}{a^2}\)

Tương tự: \(\frac{1}{2b-1}\ge\frac{1}{b^2};\frac{1}{2c-1}\ge\frac{1}{c^2}\)

Từ đó suy ra \(VT\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{4ab}{ab+1}+\frac{4bc}{bc+1}+\frac{4ca}{ca+1}\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+4-\frac{4}{ab+1}+4-\frac{4}{bc+1}+4-\frac{4}{ca+1}\)\(\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{4}{ab+1}-\frac{4}{bc+1}-\frac{4}{ca+1}+12\)\(\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}-\frac{4}{a+b}-\frac{4}{b+c}-\frac{4}{c+a}+12\)\(=\left(\frac{2}{a+b}-1\right)^2+\left(\frac{2}{b+c}-1\right)^2+\left(\frac{2}{c+a}-1\right)^2+9\ge9\)

Đẳng thức xảy ra khi a = b = c = 1

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