Trả lời hộ mình đi

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

1. Vai trò a, b, c như nhau. Không mất tính tổng quát. Giả sử \(a\ge b\ge0\)

Mà \(ab+bc+ca=3\). Do đó \(ab\ge1\)

Ta cần chứng minh rằng \(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\left(1\right)\)

Và \(\frac{2}{1+ab}+\frac{1}{1+c^2}\ge\frac{3}{2}\left(2\right)\)

Thật vậy: \(\left(1\right)\Leftrightarrow\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\ge0\\ \Leftrightarrow\left(ab-a^2\right)\left(1+b^2\right)+\left(ab-b^2\right)\left(1+a^2\right)\ge0\\ \Leftrightarrow\left(a-b\right)\left[-a\left(1+b^2\right)+b\left(1+a^2\right)\right]\ge0\\ \Leftrightarrow\left(a-b\right)^2\left(ab-1\right)\ge0\left(BĐT:đúng\right)\)

\(\left(2\right)\Leftrightarrow c^2+3-ab\ge3abc^2\\ \Leftrightarrow c^2+ca+bc\ge3abc^2\Leftrightarrow a+b+c\ge3abc\)

BĐT đúng, vì \(\left(a+b+c\right)^2>3\left(ab+bc+ca\right)=q\)

và \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\)

Nên \(a+b+c\ge3\ge3abc\)

Từ (1) và (2) ta có \(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\ge\frac{3}{2}\)

Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng BĐT Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\), ta được

\(\frac{9}{a+3b+2c}=\frac{1}{a+c+b+c+2b}\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)

Do đó ta được

\(\frac{ab}{a+3b+2c}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)=\frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)

Hoàn toàn tương tự ta được

\(\frac{bc}{2a+b+3c}\le\frac{1}{9}\left(\frac{bc}{a+b}+\frac{bc}{b+c}+\frac{b}{2}\right);\frac{ac}{3a+2b+c}\le\frac{1}{9}\left(\frac{ac}{a+b}+\frac{ac}{b+c}+\frac{c}{2}\right)\)

Cộng theo vế các BĐT trên ta được

\(\frac{ab}{a+3b+2c}+\frac{bc}{b+3c+2a}+\frac{ca}{c+3a+2b}\le\frac{1}{9}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{bc+ab}{a+c}+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}\)Vậy BĐT đc CM

ĐẲng thức xảy ra khi và chỉ khi a = b = c >0

Bài 2:

Áp dụng BĐT AM-GM:

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

\(\Rightarrow \frac{ab^2}{a^2+2b^2+c^2}\leq \frac{ab^2}{(a+b)(a+c)}\)

Hoàn toàn tương tự với các phân thức còn lại:

\(\Rightarrow \text{VT}\leq \sum \frac{ab^2}{(a+b)(a+c)}=\frac{a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)}{(a+b)(b+c)(c+a)}\)

Ta cần CM: \(\frac{a^2b^2+b^2c^2+c^2a^2+abc(a+b+c)}{(a+b)(b+c)(c+a)}\leq \frac{a+b+c}{4}\)

\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)(a+b)(b+c)(c+a)\)

\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)(a+b)(b+c)(c+a)\)

\(\Leftrightarrow 4(a^2b^2+b^2c^2+c^2a^2)+4abc(a+b+c)\leq (a+b+c)[(a+b+c)(ab+bc+ac)-abc]\)

\(\Leftrightarrow 2(a^2b^2+b^2c^2+c^2a^2)\leq (a^3b+ab^3)+(bc^3+b^3c)+(ca^3+c^3a)\)

(dễ thấy luôn đúng do theo BĐT AM-GM)

Do đó ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Đặt: f(a;b;c) =\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Vai trò của a, b, c là như nhau có thể giả sử: \(a=max\left\{a,b,c\right\}\)

Ta có: \(f\left(a;b;\sqrt{ab}\right)=\frac{a}{a+b}+\frac{b}{b+\sqrt{ab}}+\frac{\sqrt{ab}}{\sqrt{ab}+a}\)

\(=\frac{a}{a+b}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}+\frac{\sqrt{b}}{\sqrt{b}+\sqrt{a}}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

Ta chứng minh:

\(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

+) Chứng minh: \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)

Xét : \(f\left(a;b;c\right)-f\left(a;b;\sqrt{ab}\right)=\frac{b}{b+c}+\frac{c}{a+c}-\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{b\left(a+c\right)\left(\sqrt{a}+\sqrt{b}\right)+c\left(b+c\right)\left(\sqrt{a}+\sqrt{b}\right)-2\sqrt{b}\left(b+c\right)\left(a+c\right)}{\left(b+c\right)\left(a+c\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

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

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}-c\right)^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\ge0\)vì a=max{a,b,c} => \(a\ge b\)

=> \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\)(1)

+) Chứng minh:\(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

Xét: \(f\left(a;b;\sqrt{ab}\right)-\frac{7}{5}=\frac{a}{a+b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{7}{5}\)\(=\frac{\frac{a}{b}}{\frac{a}{b}+1}+\frac{2}{\sqrt{\frac{a}{b}}+1}-\frac{7}{5}\)(2)

Đặt \(\sqrt{\frac{a}{b}}=x\left(đk:x\le3\right)\)Ta có: 

(2)=\(\frac{x^2}{x^2+1}+\frac{2}{x+1}-\frac{7}{5}\)\(=\frac{5x^3+5x^2+10x^2+10-7x^3-7x^2-7x-7}{5\left(x^2+1\right)\left(x+1\right)}\)

\(=\frac{-2x^3+8x^2-7x+3}{5\left(x^2+1\right)\left(x+1\right)}=\frac{\left(3-x\right)\left(2x^2-2x+1\right)}{5\left(x^2+1\right)\left(x+1\right)}\ge0\)

=> \(f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)(3)

Từ (1); (3) => \(f\left(a;b;c\right)\ge f\left(a;b;\sqrt{ab}\right)\ge\frac{7}{5}\)

"=" xảy ra <=> a=3; b=1/3; c=1 và các hoán vị

\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)

\(\Rightarrow\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}\cdot\frac{b+c}{4bc}}=\frac{1}{a}\)

\(\Rightarrow\frac{ca}{b^2\left(c+a\right)}+\frac{c+a}{4ca}\ge2\sqrt{\frac{ca}{b^2\left(c+a\right)}\cdot\frac{c+a}{4ca}}=\frac{1}{b}\)

\(\Rightarrow\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}\cdot\frac{a+b}{4ab}}=\frac{1}{c}\)

Cộng theo vế các bất đẳng thức trên ta được:

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

Mà\(\frac{b+c}{4bc}+\frac{c+a}{4ca}+\frac{a+b}{4ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)nên:

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

hay\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)

Bất đẳng thức xảy ra khi \(a=b=c\)

bạn giỏi quá

Nguyễn Đăng Nhân

\(\Leftrightarrow\Sigma_{cyc}\frac{\left(ab+bc+ca\right)^2}{2a^2+bc}\le\left(a+b+c\right)^2\)

Ta có: \(\frac{\left(ab+bc+ca\right)^2}{2a^2+bc}\le\frac{\left(ab+ca\right)^2}{2a^2}+\frac{\left(bc\right)^2}{bc}=\frac{\left(b+c\right)^2}{2}+bc\)

Tương tự rồi cộng lại ta thu được:

\(L.H.S\le\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{2}+ab+bc+ca\)

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

P/s: Nhìn đơn giản chứ nó là bao nhiêu ngày suy nghĩ đấy ạ:( Chả biết đúng hay sai nữa:v

Bài 2:

\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)

\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)

\(\Rightarrow P\ge\sqrt[3]{3}\)

Dấu bằng xẩy ra khi a=b=c=3

Bài 1: 

 \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)

Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)

\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

Áp dụng bđt AM-GM ta có:

 \(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)

\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)

\(\Rightarrow\)(*) luôn đúng

Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)

Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)

Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)

\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)

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Làm sai rồi ạ

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)