Đáp án:
Giải thích các bước giải:
$\dfrac{a}{a+\sqrt[]{a+bc}}=\dfrac{a}{a+\sqrt[]{a(a+b+c)+bc}}\\
=\dfrac{a}{a+\sqrt[]{(a+b)(c+a)}}\leq \dfrac{a}{a+\sqrt[]{ac}+\sqrt[]{ab}}\\
=\dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}+\sqrt[]{c}}\\
\rightarrow \dfrac{a}{a+\sqrt[]{a+bc}}\leq \dfrac{\sqrt[]{a}}{\sqrt[]{a}+\sqrt[]{b}+\sqrt[]{c}}\\
\text{Tương tự: }\\
\quad \dfrac{b}{b+\sqrt[]{b+ca}}\leq \dfrac{\sqrt[]{b}}{\sqrt[]{a}+\sqrt[]{b}+\sqrt[]{c}}\\
\quad \dfrac{c}{c+\sqrt[]{c+ab}}\leq \dfrac{\sqrt[]{c}}{\sqrt[]{a}+\sqrt[]{b}+\sqrt[]{c}}\\
\Rightarrow \text{Biểu thức: } \leq 1$