Tất cả các câu này đều có thể chứng minh bằng phép biến đổi tương đương:

a.

\(\Leftrightarrow a^{10}+b^{10}+a^4b^6+a^6b^4\le2a^{10}+2b^{10}\)

\(\Leftrightarrow a^{10}-a^6b^4+b^{10}-a^4b^6\ge0\)

\(\Leftrightarrow a^6\left(a^4-b^4\right)-b^6\left(a^4-b^4\right)\ge0\)

\(\Leftrightarrow\left(a^6-b^6\right)\left(a^4-b^4\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2\right)\left(a^2+b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^2+b^2\right)\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

b.

\(\Leftrightarrow\left(\dfrac{a^2}{4}+b^2+c^2-ab+ac-2bc\right)+b^2-2b+1+c^2\ge0\)

\(\Leftrightarrow\left(\dfrac{a}{2}-b+c\right)^2+\left(b-1\right)^2+c^2\ge0\) (luôn đúng)

c.

\(\Leftrightarrow a^2+4b^2+4c^2-4ab-8bc+4ac\ge0\)

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

d.

\(\Leftrightarrow4a^4-8a^3+4a^2+a^2-2a+1\ge0\)

\(\Leftrightarrow\left(2a^2-2a\right)^2+\left(a-1\right)^2\ge0\) (luôn đúng)

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{5}{4}\left(a+b\right)^2}=\frac{\sqrt{5}\left(a+b\right)}{2}\)

Tương tự:\(\sqrt{2b^2+bc+2c^2}\ge\frac{\sqrt{5}\left(b+c\right)}{2}\);\(\sqrt{2c^2+ca+2a^2}\ge\frac{\sqrt{5}\left(c+a\right)}{2}\)

Cộng theo vế 3 BĐT trên ta có:\(VT\ge\frac{\sqrt{5}\left(2a+2b+2c\right)}{2}=\sqrt{5}\left(a+b+c\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

\(\sqrt{a^2+2ab+2b^2}=\sqrt{\left(a+b\right)^2+b^2}=\dfrac{1}{\sqrt{5}}\sqrt{\left(4+1\right)\left[\left(a+b\right)^2+b^2\right]}\ge\dfrac{1}{\sqrt{5}}\left(2a+2b+b\right)=\dfrac{1}{\sqrt{5}}\left(2a+3b\right)\)

Tương tự:

\(\sqrt{b^2+2bc+2c^2}\ge\dfrac{1}{\sqrt{5}}\left(2b+3c\right)\)

\(\sqrt{c^2+2ca+2a^2}\ge\dfrac{1}{\sqrt{5}}\left(2c+3a\right)\)

Cộng vế:

\(P\ge\dfrac{1}{\sqrt{5}}\left(5a+5b+5c\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)