Cho [tex]a,\,b,\,c>0[/tex] .Chứng minh rằng :
1) [tex]\dfrac{a^4+b^4+c^4}{ab+bc+ca}+\dfrac{abc}{a+b+c}\geq \dfrac{2}{3}\left(a^2+b^2+c^2\right)[/tex]
Giải:
Theo Bất đẳng thức [tex]Schur[/tex] bậc 1 ta có:
[tex]a^3+b^3+c^3+3abc\ge a^2\left(b+c\right)+b^2\left(c+a\right)+c^2\left(a+b\right)[/tex]
[tex]\Leftrightarrow 2\left(a^3+b^3+c^3\right)+3abc\ge\left(a^2+b^2+c^2\right)\left(a+b+c\right)[/tex]
[tex]\Leftrightarrow \dfrac{2\left(a^3+b^3+c^3\right)}{a+b+c}+\dfrac{3abc}{a+b+c}\ge a^2+b^2+c^2[/tex]
[tex]\Leftrightarrow \dfrac{4\left(a^3+b^3+c^3\right)}{3\left(a+b+c\right)}+\dfrac{2abc}{a+b+c}\ge \dfrac{2}{3}\left(a^2+b^2+c^2\right)[/tex] [tex]\left(1\right)[/tex]
Theo Bất đẳng thức [tex]Schur[/tex] bậc 2 ta có:
[tex]a^4+b^4+c^4+abc\left(a+b+c\right)\ge a^3\left(b+c\right)+b^3\left(c+a\right)+c^3\left(a+b\right)[/tex]
[tex]\Leftrightarrow 2\left(a^4+b^4+c^4\right)+abc\left(a+b+c\right)\ge\left(a^3+b^3+c^3\right)\left(a+b+c\right)[/tex]
[tex]\Leftrightarrow \dfrac{2\left(a^4+b^4+c^4\right)}{\left(a+b+c\right)^2}+\dfrac{abc}{a+b+c}\ge\dfrac{a^3+b^3+c^3}{a+b+c}[/tex] [tex]\left(2\right)[/tex]
Từ [tex]\left(1\right)[/tex] và [tex]\left(2\right)[/tex] ta có: [tex]\dfrac{\left(a^3+b^3+c^3\right)}{3\left(a+b+c\right)}+\dfrac{3abc}{a+b+c}+\dfrac{2\left(a^4+b^4+c^4\right)}{\left(a+b+c\right)^2}\ge \dfrac{2}{3}\left(a^2+b^2+c^2\right)[/tex] [tex]\left(3\right)[/tex]
Mà ta có:
[tex]\left(a^3+b^3+c^3\right)\left(a+b+c\right)\le3\left(a^4+b^4+c^4\right)[/tex]
[tex]\Rightarrow \dfrac{\left(a^3+b^3+c^3\right)}{3\left(a+b+c\right)}\le\dfrac{a^4+b^4+c^4}{\left(a+b+c\right)^2}[/tex]
[tex]\Rightarrow \dfrac{\left(a^3+b^3+c^3\right)}{3\left(a+b+c\right)}+\dfrac{2\left(a^4+b^4+c^4\right)}{\left(a+b+c\right)^2}\le\dfrac{3\left(a^4+b^4+c^4\right)}{\left(a+b+c\right)^2}\le\dfrac{a^4+b^4+c^4}{ab+bc+ca}[/tex] [tex]\left(4\right)[/tex]
Từ [tex]\left(3\right)[/tex] và [tex]\left(4\right)[/tex] ta có đpcm. [tex]\blacksquare[/tex]
