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Ta có :
\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow a^2+b^2+1-ab-a-b\ge0\)
\(\Leftrightarrow2a^2+2b^2-2ab-2a-2b+2\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) ( đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: (a-b)^2 ≥ 0
(=). a^2+b^2≥2ab
Tương tự: b^2+c^2 ≥ 2bc
c^2+a^2 ≥ 2ca
Suy ra 2×(a^2+b^2+c^2) ≥ 2×(ab+BC+ca)
(=) a^2+b^2+c^2 ≥ ab+bc+ca
Dấu bằng xảy ra khi: a=b=c
\(a^2+b^2\ge2\sqrt{a^2b^2}\ge2ab\)
\(b^2+c^2\ge2\sqrt{b^2c^2}\ge2bc\)
\(c^2+a^2\ge2\sqrt{c^2a^2}\ge2ca\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)
\(\Leftrightarrow\)a^2+2ab+b^2>=4ab
\(\Leftrightarrow\)a^2-2ab+b^2>=0
\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)
b,\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) luôn đúng
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(a-b\right)^2\ge0\)
\(\Rightarrow a^2+b^2+2ab\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
Có : \(a,b\ge0\)
\(\Rightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\) ( đpcm )
Vậy ...