A\(\ge3\)

You know

A\(\ge\)9

Theo gt \(ab+bc+ca\le abc^{\left(3\right)}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)

\(\frac{9}{a+b+c}\le1\)

\(a+b+c\ge9^{\left(1\right)}\)

Mặt khác 

\(a^2+b^2+c^2\ge3\left(a+b+c\right)\)

\(a^2+b^2+c^2\ge9\cdot3=27^{\left(2\right)}\)

Vì a,b,c >0, áp dụng bất đẳng thức cô si ta có:

\(\frac{a^2b}{b+2a}+\frac{b\left(b+2a\right)}{9}\ge2\sqrt{\frac{a^2b}{b+2a}\cdot\frac{b\left(b+2a\right)}{9}}=\frac{2ab}{3}\)

CMTT

\(\frac{b^2c}{c+2b}+\frac{c\left(c+2b\right)}{9}\ge\frac{2bc}{3}\)

\(\frac{c^2a}{a+2c}+\frac{a\left(a+2c\right)}{9}\ge\frac{2ca}{3}\)

Cộng vế với vế a được :

\(A+\frac{a^2+b^2+c^2}{9}+\frac{2\left(ab+bc+ca\right)}{9}\ge\frac{2\left(ab+bc+ca\right)}{3}\)

\(A\ge\frac{4\left(ab+bc+ca\right)}{3}-\frac{a^2+b^2+c^2}{9}^{\left(#\right)}\)

Từ 1,2,3 và # ta có

\(A\ge\frac{4\cdot9}{3}-\frac{27}{9}=9\)

Dấu bằng xảy ra \(\Leftrightarrow a=b=c=3\)

Vậy...

Ta có :\(A=\frac{1}{ab}+\frac{1}{a^2+b^2}=\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{3}{2ab}\)

           \(A\ge\frac{4}{2ab+a^2+b^2}+\frac{3}{2ab}\)

           \(A\ge\frac{4}{\left(a+b\right)^2}+\frac{3}{\frac{\left(a+b\right)^2}{2}}\)

         \(A\ge4+6=10\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a^2+b^2=2ab\\a+b=1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=b\\a+b=1\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)

Vậy Min A = 10 <=> a = b = 1/2

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,

\(A=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)

\(\ge1+1+2\sqrt{\frac{a}{b}.\frac{b}{a}}+a+b+\frac{a+b}{ab}=4+a+b+\frac{4\left(a+b\right)}{\left(a+b\right)^2}=4+a+b+\frac{4}{a+b}\)

lại có \(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a+b\le\sqrt{2}\)

\(4+a+b+\frac{4}{a+b}=4+\left(a+b+\frac{2}{a+b}\right)+\frac{2}{a+b}\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)

\(\Rightarrow A\ge4+3\sqrt{2}\)

câu 2

ta có:\(\left(2b^2+a^2\right)\left(2+1\right)\ge\left(2b+a\right)^2\Rightarrow3c\ge a+2b\)

\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\left(Q.E.D\right)\)

Vì a,b >0

Áp dụng bất đẳng thức Cauchy, ta có:

\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2.\frac{1}{a^2}}\)

                    \(\ge2\)

\(b^2+\frac{1}{b^2}\ge2\sqrt{b^2.\frac{1}{b^2}}\)

                    \(\ge2\)

Cộng vế theo vế, ta được:

\(a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\ge2+2\)

\(\Rightarrow A\ge4\)

Vậy MinA=4 \(\Leftrightarrow\orbr{\begin{cases}a^2=\frac{1}{a^2}\\b^2=\frac{1}{b^2}\end{cases}}\)

                       \(\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)

Btđ cô si

a+b=1 mà nếu a=b=1 thì a+b =2 ko được

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

\(>=\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+ac+bc}\)(bđt svacxo)\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+ac+bc}+\frac{1}{ab+ac+bc}+\frac{7}{ab+ac+bc}\)

\(>=\frac{9}{a^2+b^2+c^2+ab+ac+bc+ac+ac+bc}+\frac{7}{ab+ac+bc}\)(bđt svacxo)

\(=\frac{9}{a^2+b^2+c^2+2ab+2ac+2bc}+\frac{7}{ab+ac+bc}=\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+ac+bc}\)

\(=\frac{9}{1}+\frac{7}{ab+ac+bc}=9+\frac{7}{ab+ac+bc}\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc>=ab+ac+bc+2ab+2ac+2bc\)

\(=3ab+3ac+3bc=3\left(ab+ac+bc\right)\Rightarrow\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\cdot1=\frac{1}{3}>=ab+ac+bc\Rightarrow ab+ac+bc< =\frac{1}{3}\)

\(\Rightarrow9+\frac{7}{ab+ac+bc}>=9+\frac{7}{\frac{1}{3}}=9+7\cdot3=9+21=30\)

\(\Rightarrow A>=30\)dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

vậy min A là 30 khi \(a=b=c=\frac{1}{3}\)

1.

\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0\Leftrightarrow a^2\le}2+a\)

Tương tự \(b^2\le2+b,c^2\le2+c\Rightarrow a^2+b^2+c^2\le6+a+b+c=6\)

Dấu "=" xảy ra khi a=2,b=c=-1 và các hoán vị của chúng

Xét \(\frac{a^2+1}{a}=a+\frac{1}{a}\)

Dễ thấy dấu "=" xảy ra khi  \(a=\frac{1}{3}\)

khi đó \(a+\frac{1}{a}=a+\frac{1}{9a}+\frac{8}{9a}\ge2\sqrt{\frac{a.1}{9a}}+\frac{8}{\frac{9.1}{3}}=\frac{10}{3}\)

\(\Rightarrow\frac{a}{a^2+1}\le\frac{3}{10}\)

tương tự =>đpcm

lười quá khỏi nghĩ đưa link

| Inequalities (ko dịch dc thì pm)

Em không chắc đâu nha, sai thì xin thông cảm cho ạ

\(a=b=c=\frac{\sqrt{3}}{3}\Rightarrow B=\frac{3\sqrt{3}}{2}\). Ta se chung minh do la gia tri min cua B. That vay:

\(BĐT\Leftrightarrow\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}}{2}=\frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}\)

BĐT trên đồng bậc, nên ta chuẩn hóa a² + b² + c² = 3 và chứng minh:

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

Ta chứng minh BĐT sau: \(\frac{a}{3-a^2}\ge\frac{1}{2}a^2\Leftrightarrow\frac{a^2}{2}-\frac{a}{3-a^2}\le0\)

\(\Leftrightarrow\frac{-\left(a-1\right)^2a\left(a+2\right)}{2\left(3-a^2\right)}\le0\) (Đúng)

Tương tự với hai BĐT còn lại và cộng theo vế suy ra BĐT (2) là đúng.

Suy ra BĐT (1) là đúng suy ra \(B_{min}=\frac{3\sqrt{3}}{2}\)

Vậy...

Xét \(\frac{a}{b^2+c^2}=\frac{a}{1-a^2}\ge\frac{3\sqrt{3}}{2}a^2\)

<=> \(a^4-a^2+\frac{2\sqrt{3}}{9}a\ge0\)

<=> \(a\left(a+\frac{2\sqrt{3}}{3}\right)\left(a-\frac{\sqrt{3}}{3}\right)^2\ge0\)luôn đúng

=> \(B\ge\frac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\frac{3\sqrt{3}}{2}\)

Min \(B=\frac{3\sqrt{3}}{2}\)khi \(a=b=c=\frac{\sqrt{3}}{3}\)