Quay lại đáp án

A HUA A!!!! A HUA A!!! I'M TAZAN!!!!câu 10có $a^{2}+bc\geq 2a\sqrt{bc} \Rightarrow \frac{1}{a^{2}+bc}\leq \frac{1}{2a\sqrt{bc}}$$\Rightarrow VT\leq \frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}$ $=\frac{1}{2} \frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{abc}\leq \frac{\frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}}{2abc}$ $=\frac{a+b+c}{2abc}$ dấu "="$ \Leftrightarrow a=b=c$

A HUA A!!!! A HUA A!!! I'M TAZAN!!!!( phiên bản tazan cực dễ thương)Câu 10có $a^{2}+bc\geq 2a\sqrt{bc} \Rightarrow \frac{1}{a^{2}+bc}\leq \frac{1}{2a\sqrt{bc}}$$\Rightarrow VT\leq \frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}$ $=\frac{1}{2} \frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{abc}\leq \frac{\frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}}{2abc}$ $=\frac{a+b+c}{2abc}$ dấu "="$ \Leftrightarrow a=b=c$

câu 10có $a^{2}+bc\geq 2a\sqrt{bc} \Rightarrow \frac{1}{a^{2}+bc}\leq \frac{1}{2a\sqrt{bc}}$$\Rightarrow VT\leq \frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}$ $=\frac{1}{2} \frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{abc}\leq \frac{\frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}}{2abc}$ $=\frac{a+b+c}{2abc}$ dấu "="$ \Leftrightarrow a=b=c$

A HUA A!!!! A HUA A!!! I'M TAZAN!!!!câu 10có $a^{2}+bc\geq 2a\sqrt{bc} \Rightarrow \frac{1}{a^{2}+bc}\leq \frac{1}{2a\sqrt{bc}}$$\Rightarrow VT\leq \frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}$ $=\frac{1}{2} \frac{\sqrt{bc}+\sqrt{ca}+\sqrt{ab}}{abc}\leq \frac{\frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}}{2abc}$ $=\frac{a+b+c}{2abc}$ dấu "="$ \Leftrightarrow a=b=c$