Đặt \(\left\{{}\begin{matrix}a+b=8+x\\b=3+y\end{matrix}\right.\left(x,y\in N,xy\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=5+x-y\\b=3+y\end{matrix}\right.\)

Khi đó:

\(27a^2+10b^3=27\left(5+x-y\right)^2+10\left(3+y\right)^3\)

\(=27\left(25+x^2+y^2+10x-2xy-10y\right)+10\left(27+y^3+9y^2+27y\right)\)

\(=945+27\left(x^2+y^2-2xy\right)+270x+10y^3+90y^{2\text{​​}}\)

\(=945+27\left(x-y\right)^2+270x+10y^3+90y^2>945\)

Vậy \(27a^2+10b^3>945\)

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 \(a=b=1\)

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 \(a=b=c=1\)

\(b\left(a-b\right)\le\dfrac{\left(b+a-b\right)^2}{4}=\dfrac{a^2}{4}\)

\(\Rightarrow\dfrac{1}{b\left(a-b\right)}\ge\dfrac{4}{a^2}\)

\(\Rightarrow a+\dfrac{1}{b\left(a-b\right)}\ge a+\dfrac{4}{a^2}=\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{4}{a^2}\ge3\sqrt[3]{\dfrac{a}{2}\dfrac{a}{2}\dfrac{4}{a^2}}=3\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{a}{2}=\dfrac{4}{a^2}\\b=a-b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

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