\(\sqrt{2\left(b+c\right)^2+bc}\le\sqrt{2\left(b+c\right)^2+\frac{1}{4}\left(b+c\right)^2}=\frac{3}{2}\left(b+c\right)\)

\(\Rightarrow\frac{\left(1-c\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\frac{2}{3}.\frac{\left(1-c\right)^2}{\left(b+c\right)}\)

Tương tự ta có:

\(Q\ge\frac{2}{3}\left(\frac{\left(1-c\right)^2}{b+c}+\frac{\left(1-a\right)^2}{a+c}+\frac{\left(1-b\right)^2}{a+b}\right)\)

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

\(Q_{min}=\frac{4}{3}\) khi \(a=b=c=\frac{1}{3}\)

a )

Áp dụng BĐT Bunhiacopxki ta có :

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

\(\Rightarrow\sqrt{\frac{a^2}{b^2+\left(c+a\right)^2}}\le\sqrt{5}.\frac{a}{b+2c+2a}\)

\(\Rightarrow VT\le\sqrt{5}.\left(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\right)\)

Cần chứng minh : \(\frac{a}{b+2c+2a}+\frac{b}{c+2a+2b}+\frac{c}{a+2b+2c}\le\frac{3}{5}\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{a}{b+2c+2a}\right)+\left(\frac{1}{2}-\frac{b}{c+2a+2b}\right)+\left(\frac{1}{2}-\frac{c}{a+2b+2c}\right)\ge\frac{9}{10}\)

\(\Leftrightarrow\frac{b+2c}{b+2c+2a}+\frac{c+2a}{c+2a+2b}+\frac{a+2b}{a+2b+2c}\ge\frac{9}{5}\)

Áp dụng BĐT Bunhiacopxki dạng phân thức ở vế trái :

\(\Rightarrow VT\ge\frac{\left(b+2c+c+2a+a+2b\right)^2}{\left(b+2c\right)^2+2a\left(b+2c\right)+\left(c+2a\right)^2+2b\left(c+2a\right)+\left(a+2b\right)^2+2c\left(a+2b\right)}\)

\(=\frac{9\left(a+b+c\right)^2}{5\left(a+b+b\right)^2}=\frac{9}{5}\left(đpcm\right)\)

Dấu " = '" xảy ra khi a=b=c

b ) Ta có abc =1

Ta chứng minh :

\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}=1\)

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

\(=\frac{1}{ab+a+1}+\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}=1\left(đpcm\right)\)

Ta có : \(\left(1+a\right)^2+b^2+5=\left(a^2+b^2\right)+2a+6\ge2ab+2a+6\)

\(\Rightarrow\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}=\frac{2ab+2a+6}{ab+a+4}=2-\frac{2}{ab+a+4}\)

Mà \(\frac{1}{ab+a+4}=\frac{1}{ab+a+1+3}\le\frac{1}{4}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)\) ( do \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(\Rightarrow\frac{\left(1+a\right)^2+b^2+5}{ab+a+4}\ge2-\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{3}\right)=\frac{11}{6}-\frac{1}{2}.\frac{1}{ab+a+1}\)

Khi đó :

\(P\ge\frac{11}{2}-\frac{1}{2}.\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\right)=\frac{11}{2}-\frac{1}{2}.1=5\)

\(P_{Min}=5\) khi \(a=b=c=1\)

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, 

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

\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)

\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)

Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)

Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)

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

Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.

đúng rồi

 chó điên