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[namdang248]

cho $a,b,c$ là 3 số thực dương sao cho $ab+bc+ca=1$ chứng minh rằng

$\frac{1}{4a^{2}-bc+1}+\frac{1}{4b^{2}-ca+1}+\frac{1}{4c^{2}-ab+1}\geq \frac{3}2{}$

[Chung Anh]

cho $a,b,c$ là 3 số thực dương sao cho $ab+bc+ca=1$ chứng minh rằng

$\frac{1}{4a^{2}-bc+1}+\frac{1}{4b^{2}-ca+1}+\frac{1}{4c^{2}-ab+1}\geq \frac{3}2{}$

Đặt $x=bc;y=ac;z=ab$ => $x+y+z=1$

Khi đó $a^2=\frac{yz}{x} $

=>$VT=\frac{1}{\frac{4yz}{x}-x+1}+\frac{1}{\frac{4zx}{y}-y+1}+\frac{1}{\frac{4xy}{z}-z+1}$

        $=\frac{1}{\frac{4yz}{x}+y+z}+\frac{1}{\frac{4xz}{y}+z+x}+\frac{1}{\frac{4xy}{z}+y+x}$

        $=\frac{x^2}{3xyz+z(yz+yx+xz)}+\frac{y^2}{3xyz+y(xz+yz+yx)}+\frac{z^2}{3xyz+z(xy+yz+xz)}$

        $\geq \frac{(x+y+z)^2}{9xyz+(x+y+z)(xy+yz+zx)}=\frac{1}{9xyz+(xy+yz+zx)} $

Do $1=x+y+z\geq 3\sqrt[3]{xyz}\Leftrightarrow xyz\leq \frac{1}{27} $

      $1=(x+y+z)^2\geq 3(xy+yz+zx)\Leftrightarrow xy+yz+zx\leq \frac{1}{3} $

=>$VT \geq \frac{1}{9.\frac{1}{27}+\frac{1}{3}}=\frac{3}{2} $

Dấu bằng xảy ra <=> $x=y=z= \frac{1}{3}$ <=> $a=b=c= \frac{1}{\sqrt{3}} $