3.
Đặt vế trái là P
Ta có:
\(\left(2ab+2bc+2ca\right)P=\left(ab+ac+bc+ab+ac+bc\right)\left(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
4.
Giống câu 3:
\(3\left(ab+bc+ca\right)P=\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\left(ab+2ac+bc+2ab+ac+2bc\right)\)
\(3\left(ab+bc+ca\right)P\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow P\ge1\)
Dấu "=" xảy ra khi \(a=b=c\)
1.
\(\left(b+c+a\right)\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\ge\left(\sqrt{b}.\sqrt{\dfrac{a^2}{b}}+\sqrt{c}.\sqrt{\dfrac{b^2}{c}}+\sqrt{a}.\sqrt{\dfrac{c^2}{a}}\right)^2=\left(a+b+c\right)^2\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
\(\left(b+c+c+a+a+b\right)\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)\ge\left(a+b+c\right)^2\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
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