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

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

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

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

\(\Rightarrow\) đpcm

Xét hiệu VT - VP

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

Do a,b,c bình đẳng nên giả sử a\(\ge\)b\(\ge\)c, khi đó \(b\left(a-c\right)\)\(\ge\)0, c(b-a)\(\le\)0, a(c-b)\(\le\)0

\(a^3\ge b^3\ge c^3=>abc+a^3\ge abc+b^3\ge abc+c^3\)=>\(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}\le\dfrac{b\left(a-c\right)}{b\left(ac+b^2\right)}\)

=> VT -VP \(\le\) \(\dfrac{b\left(a-c\right)}{a\left(bc+a^2\right)}+\dfrac{c\left(b-a\right)}{b\left(ac+b^2\right)}+\dfrac{a\left(c-b\right)}{c\left(ab+c^2\right)}=\dfrac{ab-ac}{b\left(ac+b^2\right)}+\dfrac{ac-ab}{c\left(ab+c^2\right)}=\dfrac{a\left(b-c\right)}{b\left(ac+b^2\right)}-\dfrac{a\left(b-c\right)}{c\left(ab+c^2\right)}\)

mà \(\dfrac{1}{b\left(ac+b^2\right)}\le\dfrac{1}{c\left(ab+c^2\right)}\) nên VT-VP <0 đpcm

Ta viết bất đẳng thức đã cho lại thành

\(\sum\left[\dfrac{1}{c}-\dfrac{\left(a+b+2c\right)}{2\left(ab+c^2\right)}\right]\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{2\prod\left(ab+c^2\right)}\)

\(\Leftrightarrow\sum\dfrac{c\left(a^2+ab+b^2\right)\left(a-b\right)^2}{ab\left(a^2+bc\right)\left(b^2+ca\right)}\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\)

Hay \(S_a\left(b-c\right)^2+S_b\left(c-a\right)^2+S_c\left(a-b\right)^2\ge\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}\quad\left(1\right)\)

Vậy $VT\geq 0$ và $S_a+S_b\ge 0;S_b+S_c\ge 0.$ Nếu \(a\ge b\ge c\rightarrow VT\ge0\ge VP,\) ta chỉ xét \(a\le b\le c.\)

\(\left(1\right)\Leftrightarrow\left(S_a+S_b\right)\left(b-c\right)^2+\left(S_b+S_c\right)\left(a-b\right)^2\ge\left[\dfrac{\left(c-a\right)\left(a^2+b^2+c^2\right)}{\prod\left(ab+c^2\right)}-2S_b\right]\left(a-b\right)\left(b-c\right)\)

Đặt \(c=a+x+y,b=a+x\Rightarrow x=b-a;y=c-b\left(x,y\ge0\right)\) thay vào rút gọn các thứ là đpcm.

P/s: Cách này khá trâu nhưng chịu thôi, bài này mình nghĩ khá chặt.

Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:

\(VT=\dfrac{a^3+b^3+c^3}{2abc}+\dfrac{a^2+b^2}{c^2+ab}+\dfrac{b^2+c^2}{a^2+bc}+\dfrac{a^2+c^2}{b^2+ac}\ge\dfrac{3abc}{2abc}+\dfrac{2ab}{c^2+ab}+\dfrac{2bc}{a^2+bc}+\dfrac{2ac}{b^2+ac}=\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\)

Áp dụng bất đẳng thức \(Cauchy-Schwarz\) \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}=\dfrac{a^2b^2}{c^2ab+a^2b^2}+\dfrac{b^2c^2}{a^2bc+b^2c^2}+\dfrac{a^2c^2}{b^2ac+a^2c^2}\ge\dfrac{\left(ab+bc+ac\right)^2}{c^2ab+a^2b^2+a^2bc+b^2c^2+b^2ac+a^2c^2}\)

Đặt: \(\left\{{}\begin{matrix}ab=x\\bc=y\\ac=z\end{matrix}\right.\) ta được: \(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+xy+xz+xy}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)

Nên: \(\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)

Mà: \(VT\ge\dfrac{3}{2}+2\left(\dfrac{ab}{c^2+ab}+\dfrac{bc}{a^2+bc}+\dfrac{ac}{b^2+ac}\right)\Leftrightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Lời giải:

Áp dụng BĐT AM-GM ta có: \(\frac{a^3+b^3+c^3}{2abc}\geq \frac{3\sqrt[3]{a^3b^3c^3}}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\) (1)

Áp dụng BĐT Cauchy-Schwarz:

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

Có:

\((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2=2(a^2+b^2+c^2)+2\sqrt{(a^2+b^2)(b^2+c^2)}+2\sqrt{(b^2+c^2)(c^2+a^2)}+\sqrt{(a^2+b^2)(c^2+a^2)}\)

Áp dụng BĐT Bunhiacopxky:

\(\sqrt{(a^2+b^2)(b^2+c^2)}\geq \sqrt{(ac+b^2)^2}=ac+b^2\)

\(\sqrt{(b^2+c^2)(c^2+a^2)}\geq \sqrt{(ba+c^2)^2}=ba+c^2\)

\(\sqrt{(a^2+b^2)(c^2+a^2)}\geq \sqrt{(a^2+bc)^2}=a^2+bc\)

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

\(\geq a^2+b^2+c^2+ab+bc+ac+2(a^2+b^2+c^2+ab+bc+ac)\) (AM-GM)

Hay \((\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})^2\geq 3(a^2+b^2+c^2+ab+bc+ac)\) (3)

Từ \((2); (3)\Rightarrow \frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{a^2+c^2}{b^2+ac}\geq 3\) (4)

Từ \((1); (4)\Rightarrow \frac{a^3+b^3+c^3}{2abc}+\frac{a^2+b^2}{c^2+ab}+\frac{b^2+c^2}{a^2+bc}+\frac{c^2+a^2}{b^2+ac}\geq \frac{9}{2}\)

Ta có đpcm.

Dấu bằng xảy ra khi $a=b=c$

3/ Áp dụng bất đẳng thức AM-GM, ta có :

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

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

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

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

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

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

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

\(VT=\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2a\sqrt{bc}}+\dfrac{1}{2b\sqrt{ac}}+\dfrac{1}{2c\sqrt{ab}}\)

\(VT\le\dfrac{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}}{2abc}\)

Mặt khác ta luôn có:

\(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)

\(\Rightarrow2\left(a+b+c\right)-2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)\ge0\)

\(\Rightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\le a+b+c\)

\(\Rightarrow VT\le\dfrac{a+b+c}{2abc}\)

Dấu "=" khi \(a=b=c\)

Lời giải:

Áp dụng BĐT AM-GM cho các số dương:

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

Do đó:

\(\text{VT}=\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2bc}}+\frac{1}{2\sqrt{b^2ac}}+\frac{1}{2\sqrt{c^2ab}}\)

hay \(\text{VT}\leq \frac{\sqrt{bc}+\sqrt{ac}+\sqrt{ab}}{2abc}(*)\)

Tiếp tục áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} \sqrt{bc}\leq \frac{b+c}{2}\\ \sqrt{ac}\leq \frac{a+c}{2}\\ \sqrt{ab}\leq \frac{a+b}{2}\end{matrix}\right.\Rightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq a+b+c(**)\)

Từ \((*);(**)\Rightarrow \text{VT}\leq \frac{a+b+c}{2abc}\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Ta có:

\(\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{a^2bc}}=\dfrac{1}{2a\sqrt{bc}}=\dfrac{\sqrt{bc}}{2abc}\)

Tương tự:

\(\Rightarrow VT\le\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\le\dfrac{a+b+c}{2abc}\)

Dấu "=" khi a=b=c

a;b;c dương k bn