Cho a,b,c là độ dài 3 cạnh tam giác
Cm 2$a^{2}$$b^{2}$ + 2$b^{2}$$c^{2}$ +2$c^{2}$$a^{2}$ – $a^{4}$ – $b^{4}$ – $c^{4}$ >0
Cho a,b,c là độ dài 3 cạnh tam giác
Cm 2$a^{2}$$b^{2}$ + 2$b^{2}$$c^{2}$ +2$c^{2}$$a^{2}$ – $a^{4}$ – $b^{4}$ – $c^{4}$ >0
$\begin{array}{l}\text{Ta có:}\\2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\\=(a^2b^2-b^4)+(b^2c^2-c^4)+(c^2a^2-a^4)+a^2b^2+b^2c^2+c^2a^2\\=b^2(a^2-b^2)+c^2(b^2-c^2)+a^2(c^2-a^2)+a^2b^2+b^2c^2+c^2a^2\\=b^2(a-b)(a+b)+c^2(b-c)(b+c)+a^2(c-a)(c+a)+a^2b^2+b^2c^2+c^2a^2\\=-b^2(b-a)(a+b)-c^2(c-b)(b+c)-a^2(a-c)(c+a)+a^2b^2+b^2c^2+c^2a^2 \ \ (*)\\\text{Theo BĐT tam giác, ta có:}\\\begin{cases}a+b>c\\b+c>a\\c+a>b\end{cases}\leftrightarrow \begin{cases}a-c>-b\\b-a>-c\\c-b>-a\end{cases}\\\to (*)>-b^2.(-c).c-c^2.(-a).a-a^2.(-b).b+a^2b^2+b^2c^2+c^2a^2\\\to (*)>-b^2c^2-c^2a^2-a^2b^2+a^2b^2+b^2c^2+c^2a^2=0\\\to \text{đpcm}\end{array}$