Với mọi số thực x, y ta luôn có:

\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)

Do đó:

\(a^2+1\ge2a\)

\(b^2+1\ge2b\)

\(c^2+1\ge2c\)

\(a^2+b^2\ge2ab\)

\(b^2+c^2\ge2bc\)

\(c^2+a^2\ge2ca\)

Cộng vế với vế:

\(3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ca\right)\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)+3\ge12\)

\(\Leftrightarrow a^2+b^2+c^2\ge3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(P=\dfrac{a+b}{ab}+\dfrac{2}{a+b}=a+b+\dfrac{2}{a+b}\)

\(P=\dfrac{a+b}{2}+\dfrac{2}{a+b}+\dfrac{a+b}{2}\)

\(P\ge2\sqrt{\dfrac{\left(a+b\right).2}{2\left(a+b\right)}}+\dfrac{2\sqrt{ab}}{2}=3\)

Dấu "=" xảy ra khi \(a=b=1\)

\(a+\frac{1}{a-1}=\left(a-1\right)+\frac{1}{a-1}+1\ge2\sqrt{\frac{\left(a-1\right)}{a-1}}+1=2+1=3\)

Dâú "=" xay ra khi: \(a=2\)

Với mọi số thực a;b;c ta luôn có:

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}.3^2=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)