1) Áp dụng bđt \(\frac{x^2}{m}+\frac{y^2}{n}+\frac{z^2}{p}\ge\frac{\left(x+y+z\right)^2}{m+n+p}\)  :

Ta có : \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)

Easy nà!

Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\) thì xyz = 1

BĐT trở thành: \(x^2+y^2+z^2\ge x+y+z\)

Áp dụng BĐT AM-GM,ta có: \(VT+1=\left(x^2+y^2\right)+\left(z^2+1\right)\)

\(\ge2xy+2z\ge2\sqrt{2xy.2z}=4\sqrt{xyz}=4\)

Suy ra \(VT\ge3\) (1)

Lại có: \(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)

Cộng theo vế 3 BĐT: \(VT+3\ge2\left(x+y+z\right)\)

Kết hợp (1) suy ra \(2VT\ge VT+3\ge2\left(x+y+z\right)=2VP\)

Từ đây,ta có:\(2VT\ge2VP\Rightarrow VT\ge VP^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi x = y = z = 1

P/s: Không biết cách này có đúng không?

Chuyển vế qua và đặt thừa số chung,ta cần chứng minh:

\(a^2\left(\frac{1}{b+c}-\frac{1}{c+a}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0\)

\(\Leftrightarrow\frac{a^2\left(a-b\right)}{\left(b+c\right)\left(c+a\right)}+\frac{b^2\left(b-c\right)}{\left(a+c\right)\left(a+b\right)}+\frac{c^2\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}\ge0\)

\(\Leftrightarrow\frac{a^2\left(a-b\right)\left(a+b\right)+b^2\left(b-c\right)\left(b+c\right)+c^2\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

\(\Leftrightarrow\frac{a^2\left(a^2-b^2\right)+b^2\left(b^2-c^2\right)+c^2\left(c^2-a^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)

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

\(\Leftrightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)

Đặt \(\left(a^2;b^2;c^2\right)\rightarrow\left(x;y;z\right)\).Ta cần chứng minh:

\(x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (đúng)

Dấu "=" xảy ra khi x = y = z \(\Leftrightarrow a^2=b^2=c^2\Leftrightarrow a=b=c\)

\(VT\ge\dfrac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\dfrac{b^2}{\sqrt{2\left(a^2+c^2\right)}}+\dfrac{c^2}{\sqrt{2\left(a^2+b^2\right)}}\)

Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{2019}\)

\(\Rightarrow\left\{{}\begin{matrix}a^2=\dfrac{y^2+z^2-x^2}{2}\\b^2=\dfrac{x^2+z^2-y^2}{2}\\c^2=\dfrac{x^2+y^2-z^2}{2}\end{matrix}\right.\) \(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2-x^2}{x}+\dfrac{z^2+x^2-y^2}{y}+\dfrac{x^2+y^2-z^2}{z}\)

\(\Rightarrow2\sqrt{2}VT\ge\dfrac{y^2+z^2}{x}+\dfrac{z^2+x^2}{y}+\dfrac{x^2+y^2}{z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{\left(y+z\right)^2}{2x}+\dfrac{\left(z+x\right)^2}{2y}+\dfrac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\)

\(2\sqrt{2}VT\ge\dfrac{4\left(x+y+z\right)^2}{2x+2y+2z}-\left(x+y+z\right)=x+y+z=\sqrt{2019}\)

\(\Rightarrow VT\ge\dfrac{\sqrt{2019}}{2\sqrt{2}}=\sqrt{\dfrac{2019}{8}}\) (đpcm)

\(a^2\left(\frac{1}{b+c}-\frac{1}{a+c}\right)+b^2\left(\frac{1}{a+c}-\frac{1}{a+b}\right)+c^2\left(\frac{1}{a+b}-\frac{1}{b+c}\right)\ge0.\)

\(a^2\left(\frac{a-}{b+c}\frac{b}{a+c}\right)+b^2\left(\frac{b}{a+c}\frac{-c}{a+b}\right)+c^2\left(\frac{c-}{a+b}\frac{a}{b+c}\right)\ge0.\)

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

\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2.\) cái này dễ rồi .

 áp dụng BĐT Svacxơ cho 3 số lẩ luô