Chứng minh: \(a^2+b^2+1>ab+a+b\left(a\ge0,b\ge0\right)\)
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Áp dụng BĐT cosi:
\(\left(2+a+b\right)\left(a+4b+ab\right)\ge3\sqrt[3]{2ab}\cdot3\sqrt[3]{4a^2b^2}=9\sqrt[3]{8a^3b^3}=9\cdot2ab=18ab\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=b=2\\a=4b=ab\end{matrix}\right.\left(\text{vô lí}\right)\)
Vậy dấu \("="\) ko xảy ra hay \(\left(2+a+b\right)\left(a+4b+ab\right)>18ab\)
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\(\left(a+b\right)^2-4ab\ge0\)
\(\Leftrightarrow\)\(a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\)\(a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b\)
\(a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
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Vì \(a\ge0\),\(b\ge0\),\(c\ge0\),áp dụng bđt Cauchy cho 3 số dương a,b,c ta có
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ac}\)
Nhân từng vế bđt trên =>đpcm
\(\text{có:}\frac{k}{n}+\frac{n}{k}\ge2\Leftrightarrow\frac{k}{n}-2+\frac{n}{k}\ge0\Leftrightarrow\frac{k}{n}-2\sqrt{\frac{k}{n}}.\sqrt{\frac{n}{k}}+\frac{n}{k}\ge0\Leftrightarrow\left(\sqrt{\frac{k}{n}}-\sqrt{\frac{n}{k}}\right)^2\ge0\forall k,n>0\)
\(\left(a+b\right).\left(b+c\right).\left(c+a\right)\ge8abc\)
\(\Leftrightarrow\left(ab+ac+b^2+bc\right).\left(a+c\right)\ge8abc\)
\(\Leftrightarrow a^2b+a^2c+ab^2+abc+abc+ac^2+b^2c+bc^2\ge8abc\)
\(\Leftrightarrow2+\frac{a}{c}+\frac{a}{b}+\frac{b}{c}+\frac{c}{b}+\frac{b}{a}+\frac{c}{a}\ge8\)
\(\Leftrightarrow2+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)\ge8\)(luôn đúng với mọi a,b,c >=0)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left[a+b+2\sqrt{ab}\right]^4\)
áp dụng BDT AM-GM
\(=>\left[a+b+2\sqrt{ab}\right]^4\ge\left[2\sqrt{\left(a+b\right)\left(2\sqrt{ab}\right)}\right]^4=64ab\left(a+b\right)^2\)