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[supernatural1]

Cho ba số thực không âm a,b,c thỏa mãn ab+bc+ca=1. Chứng minh rằng:

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

[phamngochung9a]

Cho ba số thực không âm a,b,c thỏa mãn ab+bc+ca=1. Chứng minh rằng:

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

$VT=\frac{1+3\left ( 1-ca-cb \right )}{a+b}+\frac{1+3\left ( 1-ab-ca \right )}{b+c}+\frac{1+3\left ( 1-ab-bc \right )}{c+a}\\=4\left ( \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right )-3\left ( a+b+c \right )\\=\frac{4\left ( a^{2}+b^{2}+c^{2}+3 \right )}{\left ( a+b+c \right )\left ( ab+bc+ca \right )-abc}-3\left ( a+b+c \right )\\\geq\frac{4\left ( a+b+c \right )^{2}+4}{a+b+c}-3\left ( a+b+c \right )=a+b+c+\frac{4}{a+b+c}\geq 4=VP\\\rightarrow Q.E.D$