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

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

\(A\ge3\left(a+b+c\right)+\frac{9}{a+b+c}=3.3+\frac{9}{3}=12\)

\(A_{min}=12\) khi \(a=b=c=1\)

 Ta cần chứng minh: \(3a+\frac{1}{a}\ge2a+2\Leftrightarrow3a+\frac{1}{a}-4\ge2\left(a-1\right)\)

\(\Leftrightarrow\frac{3a^2-4a+1}{a}-2\left(a-1\right)\ge0\Leftrightarrow\left(a-1\right)\left(\frac{3a-1}{a}-2\right)\ge0\Leftrightarrow\frac{\left(a-1\right)^2}{a}\)(đúng)

Tương tự: \(3b+\frac{1}{b}\ge2b+2;3c+\frac{1}{c}\ge2c+2\)

Cộng theo vế: \(A\ge2\left(a+b+c\right)+6=12\)

Dấu bằng xảy ra khi a=b=c=1

\(A=3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3^2+\frac{9}{a+b+c}=12.\)(BĐT Bunhia)

Dấu "=" xra khi a=b=c=1

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

\(>=\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+ac+bc}\)(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 \(a=b=c=\frac{1}{3}\)

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 \(a=b=c=\frac{\sqrt{3}}{3}\)

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

tương tự :

\(b+ac=\left(b+a\right)\left(b+c\right);c+ba=\left(b+c\right)\left(c+a\right)\)

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

áp dụng bất đẳng thức cauchy cho hai số dương 

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

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

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

cộng vế theo vế

\(P\le1\)

\(P\le\frac{3}{2}\)