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Áp dụng bđt Cauchy Schwarz dưới dạng Engel ta có :
\(\frac{\left(a+b\right)^2}{c}+\frac{\left(c+b\right)^2}{a}+\frac{\left(a+c\right)^2}{b}\ge\frac{\left(a+b+c+b+c+a\right)^2}{a+b+c}\)
\(=\frac{\left(2a+2b+2c\right)^2}{a+b+c}=\frac{4\left(a+b+c\right)^2}{a+b+c}=4\left(a+b+c\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
Xí trước phần b
Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2b+ca^2}+\frac{ca}{b^2c+ab^2}+\frac{ab}{c^2a+bc^2}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2bc^2}+\frac{c^2a^2}{ab^2c^2+a^2b^2c}+\frac{a^2b^2}{a^2bc^2+ab^2c^2}\)
\(=\frac{\left(bc\right)^2}{ab+ca}+\frac{\left(ca\right)^2}{bc+ab}+\frac{\left(ab\right)^2}{ca+bc}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
Cách làm khác của phần b ngắn gọn hơn:)
Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{\frac{1}{a^2}}{a\left(b+c\right)}+\frac{\frac{1}{b^2}}{b\left(c+a\right)}+\frac{\frac{1}{c^2}}{c\left(a+b\right)}\)
\(=\frac{\left(\frac{1}{a}\right)^2}{ab+ca}+\frac{\left(\frac{1}{b}\right)^2}{bc+ab}+\frac{\left(\frac{1}{c}\right)^2}{ca+bc}\)
\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
1)Áp dụng Bđt Am-Gm \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}=2\)
2)Áp dụng Am-Gm \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;b^2+c^2\ge2bc;a^2+c^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
=>ĐPcm
3)(a+b+c)2\(\ge\)3(ab+bc+ca)
=>a2+b2+c2+2ab+2bc+2ca\(\ge\)3ab+3bc+3ca
=>a2+b2+c2-ab-bc-ca\(\ge\)0
=>2a2+2b2+2c2-2ab-2bc-2ca\(\ge\)0
=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2)\(\ge\)0
=>(a-b)2+(b-c)2+(c-a)2\(\ge\)0
4)đề đúng \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
\(\frac{a^3}{b\left(b+c\right)}+\frac{b}{2}+\frac{b+c}{4}\ge3\sqrt[3]{\frac{a^3}{b\left(b+c\right)}.\frac{b}{2}.\frac{b+c}{4}}=\frac{3}{2}a\)
\(\Leftrightarrow\)\(\frac{a^3}{b\left(b+c\right)}\ge\frac{3}{2}a-\frac{1}{2}b-\frac{1}{4}\left(b+c\right)=\frac{3}{2}a-\frac{3}{4}b-\frac{1}{4}c\)
Tương tự, ta có: \(\frac{b^3}{c\left(c+a\right)}\ge\frac{3}{2}b-\frac{3}{4}c-\frac{1}{4}a;\frac{c^3}{a\left(a+b\right)}\ge\frac{3}{2}c-\frac{3}{4}a-\frac{1}{4}b\)
Cộng theo vế 3 bđt ta được đpcm
\(P=\frac{a^2}{b^2+2bc}+\frac{b^2}{c^2+2ac}+\frac{c^2}{a^2+2ab}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)
Dấu "=" xảy ra khi \(a=b=c\)