\(M=-x^2+12x+8=-\left(x-6\right)^2+44\le44\)

\(M_{max}=44\) khi \(x=6\)

\(N=a^2+9b^2+5a-6b=\left(a+\dfrac{5}{2}\right)^2+\left(3b-1\right)^2-\dfrac{41}{4}\ge-\dfrac{41}{4}\)

\(N_{min}=-\dfrac{41}{4}\) khi \(\left(a;b\right)=\left(-\dfrac{5}{2};\dfrac{1}{3}\right)\)

\(Q=3\left(a-5\right)^2-82\ge-82\)

\(Q_{min}=-82\) khi \(a=5\)

Mình nghĩ đề bị sai bạn ạ, bạn xem lại giùm mình nhé

\(A=\dfrac{a^2+a-1}{a^2+a+1}\)

\(\Leftrightarrow\left(A-1\right)a^2+\left(A-1\right)a+A+1=0\)

Để PT theo nghiệm a có nghiệm thì

\(\Delta=\left(A-1\right)^2-4\left(A-1\right)\left(A+1\right)\ge0\)

\(\Leftrightarrow-3A^2-2A+5\ge0\)

\(\Leftrightarrow-\dfrac{5}{3}\le A\le1\)

Vậy \(\left\{{}\begin{matrix}max=1\\min=-\dfrac{5}{3}\end{matrix}\right.\)

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

Tương tự, ta có:

\(M\ge\dfrac{\sqrt{3}}{2}\left(a+b\right)+\dfrac{\sqrt{3}}{2}\left(b+c\right)+\dfrac{\sqrt{3}}{2}\left(c+a\right)=\sqrt{3}\left(a+b+c\right)=3\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta sẽ chứng minh BĐT sau: a^2+b^2+c^2>=ab+ac+bc với mọi a,b,c

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

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

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

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

a: ab+ac+bc>=3

mà a^2+b^2+c^2>=ab+ac+bc(CMT)

nên a^2+b^2+c^2>=3

Dấu = xảy ra khi a=b=c=1

Khi a=b=c=1 thì A=1+1+1+10=13

b: a^2+b^2+c^2<=8

Dấu = xảy ra khi \(a^2=b^2=c^2=\dfrac{8}{3}\)

=>\(a=b=c=\dfrac{2\sqrt{2}}{\sqrt{3}}=\dfrac{2\sqrt{6}}{3}\)

Khi \(a=b=c=\dfrac{2\sqrt{6}}{3}\) thì \(B=\dfrac{2\sqrt{6}}{3}\cdot3-5=2\sqrt{6}-5\)

\(\)đặt \(2x^2+y^2+\dfrac{28}{x}+\dfrac{1}{y}=A\)

\(=>A=2x^2+y^2-7x-y+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

\(A=2x^2-8x+8+y^2-2y+1+x+y-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

\(A=2\left(x-2\right)^2+\left(y-1\right)^2+\left(x+y\right)-9+\dfrac{28}{x}+7x+\dfrac{1}{y}+y\)

áp dụng BDT AM-GM\(=>\dfrac{28}{x}+7x+\dfrac{1}{y}+y\ge2\sqrt{28.7}+2\sqrt{1}=30\)

\(=>A\ge30+3-9=24\)

dấu"=" xảy ra<=>x=2,y=1

\(D=\dfrac{-x+3+1}{x-3}=-1+\dfrac{1}{x-3}\)

D min khi x-3=-1

=>x=2

a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)