Đề là 

Cho \(a;b;c\ge0\) thỏa mãn a+b+c = 1

Cmr : \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\ge\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}\) ak bạn 

Ta có:a+b+c=1

\(đpcm\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{2}{a+2b+c}+\frac{2}{2a+b+c}+\frac{2}{a+b+2c}\)(*)

Áp dụng BĐT Bunhiacopxki:

\(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)(1)

Tương tự:\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{a+b+2c}\)(2)

\(\frac{1}{a+b}+\frac{1}{c+a}\ge\frac{4}{2a+b+c}\)(3)

Cộng theo từng vế của (1);(2);(3) ta đc:(*)(đpcm)

Dấu ''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)

Áp dụng BĐT AM-GM ta có:

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

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\ge a+b+c+3-\frac{a+b+c+ab+bc+ac}{2}\)

\(\ge a+b+c+3-\frac{a+b+c+\frac{\left(a+b+c\right)^2}{3}}{2}\)

\(\ge3+3-\frac{3+\frac{3^2}{3}}{2}=3\)

\("="\Leftrightarrow a=b=c=1\)

Từ giả thiết  a ≤ 1 , b ≤ 1 , c ≤ 1 ta có  a 4 ≤ a 2 , b 6 ≤ b 2 , c 8 ≤ c 2 . Từ đó  a 4 + b 6 + c 8 ≤ a 2 + b 2 + c 2

Lại có:  a − 1 b − 1 c − 1 ≤ 0   v à   a + 1 b + 1 c + 1 ≥ 0 nên

a + 1 b + 1 c + 1 − a − 1 b − 1 c − 1 ≥ 0 ⇔ 2 a b + 2 b c + 2 c a + 2 ≥ 0 ⇔ − 2 a b + b c + c a ≤ 2

Hơn nữa  a + b + c = 0 ⇔ a 2 + b 2 + c 2 = − a b + b c + c a ≤ 2

⇒ a 4 + b 6 + c 8 ≤ 2

Lời giải:

Áp dụng BĐT Cô-si: 

$a+b+c\geq 3\sqrt[3]{abc}=3(1)$
Tiếp tục áp dụng BĐT Cô-si:

$a^3+a\geq 2a^2$

$b^3+b\geq 2b^2$

$c^3+c\geq 2c^2$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)$

Lại có:

$a^2+1\geq 2a$

$b^2+1\geq 2b$

$c^2+1\geq 2c$

$\Rightarrow a^2+b^2+c^2\geq 2(a+b+c)-3=(a+b+c)+(a+b+c)-3$

$\geq a+b+c+3-3=a+b+c(2)$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)\geq a^2+b^2+c^2(3)$

Từ $(1); (2); (3)$ ta có đpcm.

Ta có: \(a+1-\frac{a+1}{b^2+1}=\frac{ab^2+b^2}{b^2+1}\le\frac{ab^2+b^2}{2b}=\frac{ab}{2}+\frac{b}{2}\) vì \(b^2+1\ge2b\)

nên \(\frac{a+1}{b^2+1}\ge a+1-\frac{b}{2}-\frac{ab}{2}\) Tương tự: 

Vậy ta có: \(VT\ge a+b+c+3-\frac{a+b+c}{2}-\frac{1}{2}\left(ab+bc+ca\right)\)

Vì \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{9}{3}=3\)

nên \(VT\ge3+\frac{a+b+c}{2}-\frac{1}{2}3=3+\frac{3}{2}-\frac{3}{2}=3=VP\)

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

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

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

Dễ dàng chứng minh \(\dfrac{b-2}{2b},\dfrac{c-2}{2c}\) là các số dương.

Áp dụng BĐT Cauchy cho 2 số dương ta có:

\(\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\ge2\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{4bc}}=\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\)

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

CMTT ta có: \(\left\{{}\begin{matrix}\dfrac{1}{b}\ge\sqrt{\dfrac{\left(c-2\right)\left(a-2\right)}{ca}}\left(2\right)\\\dfrac{1}{c}\ge\sqrt{\dfrac{\left(a-2\right)\left(b-2\right)}{ab}}\left(3\right)\end{matrix}\right.\)

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

\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\left(đpcm\right)\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\Leftrightarrow a=b=c=3\)

Đồng thời chỉ ra phương pháp nhé!!

Bunhiacopxki: \(\left(a^2+b+c\right)\left(1+b+c\right)\ge\left(a+b+c\right)^2=9\)

\(\Rightarrow\dfrac{1}{a^2+b+c}\le\dfrac{1+b+c}{9}\)

\(\Rightarrow\dfrac{a}{a^2+b+c}\le\dfrac{a+ab+ac}{9}\)

Tương tự: \(\dfrac{b}{b^2+a+c}\le\dfrac{b+ab+bc}{9}\) ; \(\dfrac{c}{c^2+a+b}\le\dfrac{c+ac+bc}{9}\)

Cộng vế:

\(P\le\dfrac{a+b+c+2\left(ab+bc+ca\right)}{9}\le\dfrac{a+b+c+\dfrac{2}{3}\left(a+b+c\right)^2}{9}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)