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gt $\Leftrightarrow \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$ (1)đặt $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}$ ($x,y,z>0$ ) (1) $\Leftrightarrow xy+yz+zx=1$VT of đpcm TT$\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{y^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{z^{2}}}}$ ta có $\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}=\frac{1}{\sqrt{\frac{x^{2}+1}{x^{2}}}}=\frac{x}{\sqrt{(x+y)(x+z)}}$ ( do $ab+bc+ca=1$)$\leq \frac{1}{2}(\frac{x}{x+y}+\frac{x}{x+z})$TT$\Rightarrow VT\leq \frac{3}{2}$dấu "="$\Leftrightarrow a=b=c=\sqrt{3}$

gt $\Leftrightarrow \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$ (1)đặt $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}$ ($x,y,z>0$ ) (1) $\Leftrightarrow xy+yz+zx=1$VT of đpcm TT$\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{y^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{z^{2}}}}$ ta có $\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}=\frac{1}{\sqrt{\frac{x^{2}+1}{x^{2}}}}=\frac{x}{\sqrt{(x+y)(x+z)}}$ ( do $xy+yz+zx=1$)$\leq \frac{1}{2}(\frac{x}{x+y}+\frac{x}{x+z})$TT$\Rightarrow VT\leq \frac{3}{2}$dấu "="$\Leftrightarrow a=b=c=\sqrt{3}$

gt $\Leftrightarrow \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$ (1)đặt $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}$ ($x,y,z>0$ ) (1) $\Leftrightarrow xy+yz+zx=1$VT of đpcm TT$\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{y^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{z^{2}}}}$ ta có $\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}=\frac{1}{\sqrt{\frac{x^{2}+1}{x^{2}}}}=\frac{x}{\sqrt{(x+y)(x+z)}}$$\leq \frac{1}{2}(\frac{x}{x+y}+\frac{x}{x+z})$TT $\Rightarrow VT\leq \frac{3}{2}$dấu "=" $\Leftrightarrow a=b=c=\sqrt{3}$

gt $\Leftrightarrow \frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}=1$ (1)đặt $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}$ ($x,y,z>0$ ) (1) $\Leftrightarrow xy+yz+zx=1$VT of đpcm TT$\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{y^{2}}}}+\frac{1}{\sqrt{1+\frac{1}{z^{2}}}}$ ta có $\frac{1}{\sqrt{1+\frac{1}{x^{2}}}}=\frac{1}{\sqrt{\frac{x^{2}+1}{x^{2}}}}=\frac{x}{\sqrt{(x+y)(x+z)}}$ ( do $ab+bc+ca=1$)$\leq \frac{1}{2}(\frac{x}{x+y}+\frac{x}{x+z})$TT $\Rightarrow VT\leq \frac{3}{2}$dấu "=" $\Leftrightarrow a=b=c=\sqrt{3}$