\(\sqrt{a}+\sqrt{b}+\sqrt{c}\ge ab+bc+ca\)
\(\Leftrightarrow2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\left(a+b+c\right)^2=3\left(a+b+c\right)\)
Ap dung BDT AM-GM ta co:
\(a^2+\sqrt{a}+\sqrt{a}\ge3a\)
\(b^2+\sqrt{b}+\sqrt{b}\ge3b\)
\(c^2+\sqrt{c}+\sqrt{c}\ge3c\)
Cong theo ve ta co DPCM
Dau "=" xay ra khi \(a=b=c=1\)
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