Áp dụng bđt Cauchy-schwarz dạng engel ta có:

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

Dấu "=" \(\Leftrightarrow\frac{a}{a+2b}=\frac{b}{b+2c}=\frac{c}{c+2a}\Leftrightarrow a=b=c\)

2. \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{\left(2a+3b\right)+\left(2b+3c\right)+\left(2c+3a\right)}=\frac{a+b+c}{5}\)

Dấu "=" \(\Leftrightarrow a=b=c\)

Áp dụng Cauchy Schwarz dạng Engel ta có :

\(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{\left(a+b+c\right)^2}{5\left(a+b+c\right)}=\frac{a+b+c}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

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

\(P\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ca\right)}\ge\frac{3\left(ab+bc+ca\right)}{5\left(ab+bc+ca\right)}=\frac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

a^3/b +a^3/b +b^2 >=3.a^2 
=>2a^3/b +b^2>=3a^2 
tuong tu 
2b^3/c +c^2 >=3.b^2 
2c^3/a +a^2 >=3.c^2 
cog lai ta dc 
2(a^3/b+b^3/c+c^3/a) +(a^2+b^2+c^2) >=3.(a^2+b^2+c^2) 
=>a^3/b+b^3/c+c^3/a >=a^2+b^2+c^2 
mat khc 
a^2+b^2+c^2>=ab+bc+ca 
nen 
a^3/b+b^3/c+c^3/a >=ab+bc+ca 
dau = xay ra khi a=b=c

k nha

a^3/b +a^3/b +b^2 >=3.a^2 
=>2a^3/b +b^2>=3a^2 
tuong tu 
2b^3/c +c^2 >=3.b^2 
2c^3/a +a^2 >=3.c^2 
cog lai ta dc 
2(a^3/b+b^3/c+c^3/a) +(a^2+b^2+c^2) >=3.(a^2+b^2+c^2) 
=>a^3/b+b^3/c+c^3/a >=a^2+b^2+c^2 
mat khc 
a^2+b^2+c^2>=ab+bc+ca 
nen 
a^3/b+b^3/c+c^3/a >=ab+bc+ca 
dau = xay ra khi a=b=c

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)

\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)

\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)

*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)

Làm lại lun ._.

a) Áp dụng bất đẳng thức AM-GM : 

\(\left(a^2+b^2\right)\left(a^2+1\right)\ge2\sqrt{a^2b^2}.2\sqrt{a^2}\ge2ab.2a=4a^2b\)

b) Áp dụng bất đẳng thức :\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\forall x;y>0\)

 \(\frac{1}{a+3b}+\frac{1}{b+2c+a}\ge\frac{4}{a+3b+b+2c+a}=\frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)

Tương tự \(\hept{\begin{cases}\frac{1}{b+3c}+\frac{1}{c+2a+b}\ge\frac{2}{b+2c+a}\\\frac{1}{c+3a}+\frac{1}{a+2b+c}\ge\frac{2}{b+2a+c}\end{cases}}\)

Cộng vế với vế ta được : \(VT+VP\ge2VP\Rightarrow VT\ge VP\)(đpcm)