a: VT=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)

\(=(x^2-ax-bx+ab)+(x^2-bx-cx+bc)+(x^2-cx-ax+ca)\)

\(=3x^2-2ax-2bx-2cx+ab+bc+ca\)

\(=3x^2-2x(a+b+c)+ab+bc+ca\)

\(=3x^2-2x\cdot2x+ab+bc+ca=-x^2+ab+bc+ca\)

=VP

b: \(2bc+b^2+c^2-a^2\)

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

\(=2p\left(b+c+a-2a\right)=2p\left(2p-2a\right)=4p\left(p-a\right)\)

BĐT cần chứng minh tương đương:

\(a^2+b^2+c^2\ge2ab-2bc+2ca\)

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

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

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

Vậy BĐT đã cho đúng

#)Giải :

Ta có : \(a+b+c=2p\)

\(\Rightarrow b+c=2p-a\)

\(\Rightarrow\left(b+c\right)^2=\left(2p-a\right)^2\)

\(\Rightarrow b^2+c^2+2bc=4p^2-4pa+a^2\)

\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

\(\Rightarrowđpcm\)

Vì a,b,c là 3 cạnh tam giác nên \(a+b>c\Leftrightarrow ac+bc>c^2\)

CMTT: \(ab+bc>b^2;ab+ac>a^2\)

Cộng vế theo vế \(\Leftrightarrow a^2+b^2+c^2< ab+bc+ca+ab+bc+ca\)

\(\Leftrightarrow a^2+b^2+c^2< 2ab+2bc+2ca\\ \Leftrightarrow a^2+b^2+c^2-2ab-2bc-2ca< 0\)