a: \(x^2-8x+5\)

\(=x^2-8x+16-11\)

\(=\left(x-4\right)^2-11\ge-11\forall x\)

Dấu '=' xảy ra khi x-4=0

=>x=4

b: \(a^3+b^3+c^3=3bac\)

=>\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

=>\(\left(a+b+c\right)\left\lbrack\left(a+b\right)^2-c\left(a+b\right)+c^2\right\rbrack-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

=>\(a^2+b^2+c^2-ab-ac-bc=0\)

=>\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)

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

=>a=b=c

\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)

\(=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3}{3^2}=\frac13\)

\(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{1}{2}ab\)

Tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{1}{2}bc\) ; \(\dfrac{c}{1+a^2}\ge c-\dfrac{1}{2}ca\)

Cộng vế:

\(P\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{6}\left(a+b+c\right)^2=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)

a) Áp dụng bất đẳng thức Bnhiacopxki ta có :

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a.1+b.1+c.1\right)^2\)

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

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

b) Ta có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)

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

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac+bc\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3ab+3bc+3ac\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)

\(\Rightarrow ab+ac+bc\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

lỗi r bn ơi

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