Áp dụng BĐT Cauchy cho 2 số dương ta được :

\(\dfrac{a^2}{b+3c}+\dfrac{b+3c}{16}\ge2\sqrt{\dfrac{a^2}{b+3c}\times\dfrac{b+3c}{16}}=\dfrac{2a}{4}\)

Suy ra \(\dfrac{a^2}{b+3c}\ge\dfrac{2a}{4}-\dfrac{b+3c}{16}\)

Cmtt ta cũng được :

\(\dfrac{b^2}{c+3a}\ge\dfrac{2b}{4}-\dfrac{c+3a}{16}\) \(\dfrac{c^2}{a+3b}\ge\dfrac{2c}{4}-\dfrac{a+3b}{16}\)

Khi đó :

\(\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{2a}{4}-\dfrac{b+3c}{16}+\dfrac{2b}{4}-\dfrac{c+3a}{16}+\dfrac{2c}{4}-\dfrac{a+3b}{16}\)

mà \(\dfrac{2a}{4}-\dfrac{b+3c}{16}+\dfrac{2b}{4}-\dfrac{c+3a}{16}+\dfrac{2c}{4}-\dfrac{a+3b}{16}=\dfrac{a+b+c}{4}\)

Vậy \(\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{a+b+c}{4}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow\dfrac{a^2}{b+3c}+\dfrac{b^2}{c+3a}+\dfrac{c^2}{a+3b}\ge\dfrac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\dfrac{a+b+c}{4}\) (đpcm)

Dấu " = " xảy ra khi \(a=b=c\)

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

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 \(\Leftrightarrow a=b=c\)

Bài 2:

Áp dụng Bdt Cauchy-Schwarz dạng engel, ta có

\(VT\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\)

Mà theo Bđt cosi 

\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\)

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

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

Dấu "=" xảy ra khi \(a=b=c\)

Đề sai

cho e lick cái đề đó luôn nà  ... ấn vào đấy    Đáp án đề thi vào lớp 10 Thanh Hóa năm 2013 | Điểm Thi Lớp 10 Thanh Hóa 2016

Áp dụng bất đẳng thức Cauchy-Schwarz ta có :

\(VT\ge\frac{\left(a+b+c\right)^2}{4\left(a+b+c\right)}=\frac{a+b+c}{4}\)

Dấu đẳng thức xảy ra \(\Leftrightarrow a=b=c\)