Đáp án:
Giải thích các bước giải:
câu 1:
$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\\
=\dfrac{a^{2}}{ab+ac}+\dfrac{b^{2}}{bc+ba}+\dfrac{c^{2}}{ca+cb}\\
=(\dfrac{a^{2}}{ab+ac}+\dfrac{b^{2}}{bc+ba}+\dfrac{c^{2}}{ca+cb})((ab+ac)+(bc+ba)+(ca+cb))\\
\geq (\sqrt[]{\frac{a^{2}}{ab+ac}.(ab+ac)}+\sqrt[]{\frac{b^{2}}{bc+ba}.(bc+ba)}+\sqrt[]{\frac{c^{2}}{ca+cb}.(ca+cb)})^{2}\\
=(a+b+c)^{2}\geq 3(ab+bc+ca)\\
\rightarrow \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq \dfrac{3(ab+bc+ca)}{(ab+ac)+(bc+ba)+(ca+cb)}=\dfrac{3}{2}$
Câu 2:
$x+y\geq 2\sqrt[]{xy}\\
y+z\geq 2\sqrt[]{yz}\\
z+x\geq 2\sqrt[]{xz}\\
\rightarrow (x+y)(y+z)(z+x)\geq 2\sqrt[]{xy}.2\sqrt[]{yz}.2\sqrt[]{xz}=8xyz$
