a.

Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b+c}{4}\ge2\sqrt{\dfrac{a^2\left(b+c\right)}{4\left(b+c\right)}}=a\)

Tương tự: \(\dfrac{b^2}{c+a}+\dfrac{c+a}{4}\ge b\) ; \(\dfrac{c^2}{a+b}+\dfrac{a+b}{4}\ge c\)

Cộng vế:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{a+b+c}{2}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

b.

Ta có:

\(a^2+bc\ge2\sqrt{a^2bc}=2\sqrt{ab.ac}\Rightarrow\dfrac{1}{a^2+bc}\le\dfrac{1}{2\sqrt{ab.ac}}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{ac}\right)\)

Tương tự: \(\dfrac{1}{b^2+ac}\le\dfrac{1}{4}\left(\dfrac{1}{ab}+\dfrac{1}{bc}\right)\) ; \(\dfrac{1}{c^2+ab}\le\dfrac{1}{4}\left(\dfrac{1}{ac}+\dfrac{1}{bc}\right)\)

Cộng vế với vế:

\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\le\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{a+b+c}{2abc}\)

Dấu "=" xảy ra khi \(a=b=c\)

Theo định lí cos ta có: \({a^2} = {b^2} + {c^2} - 2bc\;\cos A\)

\( \Rightarrow {b^2} + {c^2} - {a^2} = 2bc\;\cos A\)(1)

a) Nếu góc A nhọn thì \(\cos A > 0\)

Từ (1), suy ra \({b^2} + {c^2} > {a^2}\)

b) Nếu góc A tù thì \(\cos A < 0\)

Từ (1), suy ra \({b^2} + {c^2} < {a^2}\)

c) Nếu góc A vuông thì \(\cos A = 0\)

Từ (1), suy ra \({b^2} + {c^2} = {a^2}\)

\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\Leftrightarrow\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}\le1\)

\(\Rightarrow1\ge\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

Ta có: a^2 + b^2 + c^2 = ab + bc + ca 
<=> 2.a^2 + 2.b^2 + 2.c^2 = 2.ab + 2.bc + 2.ca 
<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc +c^2 ) + ( c^2 - 2ac + a^2 ) =0 
<=> (a-b)^2 + (b-c)^2 + (c -a)^2 =0 (1) 
Vì (a-b)^2 ; (b-c)^2 ; (c -a)^2 ≧ 0 với mọi a,b,c. 
=> (a-b)^2 + (b-c)^2 + (c -a)^2 ≧ 0 (2) 
Từ (1) và (2) khẳng định dấu "=" khi: 
a - b = 0; b - c = 0 ; c - a = 0 => a=b=c 
Vậy a=b=c.

Ta có :

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

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)

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

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

\(\Leftrightarrow\left(a^a-2ab+b^2\right)+\left(b^2-2bc+b^2\right)+\left(a^2-2bc+c^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2=0\)

Hoặc \(\Rightarrow\left(a-b\right)^2=0\) hoặc \(\left(b-c\right)^2\)hoặc \(\left(a-c\right)^2=0\Rightarrow a-b=0\)hoặc \(b-c=0\)hoặc \(a-c=0\)hoặc \(a=b\)hoặc \(b=c\)hoặc \(a=c\)

\(\Rightarrow a=b=c\left(đpcm\right)\)