c) theo bđt cauchy ta có

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+1\ge2b\\a^2+1\ge2a\end{matrix}\right.\)

cộng hết lại rút 2 đi \(\Rightarrowđpcm\)

b)theo bđt bunhiacopxki ta có

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

\(\Rightarrowđpcm\)

theo bđt cauchy ta có

\(-\left(a^2d^2+b^2c^2\right)\le-2abcd\)

\(\Leftrightarrow a^2c^2-a^2d^2+b^2d^2-b^2c^2\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow a^2(c^2-d^2)-b^2(c^2-d^2)\le a^2c^2-2abcd+b^2d^2\)

\(\Leftrightarrow(c^2-d^2)\left(a^2-b^2\right)\le(ac-bd)^2\)

\(\Rightarrowđpcm\)

Áp dụng BĐT AM - GM cho các cặp số không âm, ta được:

\(a^2+b^2\ge2ab\)(1)

\(a^2+1\ge2a\)(2)

\(b^2+1\ge2b\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được:

\(2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)

\(\Leftrightarrow a^2+b^2+1\ge ab+a+b\left(q.e.d\right)\)

Đẳng thức xảy ra khi a = b = 1

\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

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

=> đpcm

a,Ta có:\(a^2+b^2\ge2ab\)

            \(a^2+c^2\ge2ac\)  

            \(b^2+c^2\ge2bc\)

Cộng theo từng về 3 bđt trên ta đc:

\(2\left(a^2+b^2+c^2\right)\ge2\left(ab+ac+bc\right)\)

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

Xảy ra dấu đt khi \(a=b=c\)

b,\(a^3+b^3\ge ab\left(a+b\right)\)(chia cả 2 vế cho \(a+b>0\))

\(\Leftrightarrow a^2-ab+b^2\ge ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\forall a,b\)

Xảy ra dấu đẳng thức khi \(a=b\)

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

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)

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

Xảy ra đẳng thức khi \(a=b=c=0\)

Phần b mình tặng thêm một cách giải không dùng biến đổi tương đương: 

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)

Dấu bằng tại a=b

=>(ab-1)^2+ab(a-b)^2>=0

=>a^2b^2-2ab+1+ab(a^2-2ab+b^2)>=0

=>a^2b^2-2ab+1+a^3b-2a^2b^2+ab^3>=0

=>a^3b+ab^3-a^2b^2-2ab+1>=0

=>ab(a^2+b^2)-2ab-a^2b^2+1>=0

=>ab(a^2+b^2-2-ab)+1>=0(luôn đúng)

Với \(a=b=1\) BĐT sai

\(1.\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2+3\ge2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(2.\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\frac{a^2}{4}\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+d^2\ge a\left(b+c+d\right)\)
Dấu "=" xảy ra khi \(a=b=c=0\)
\(3.\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\left(\frac{a}{2}-e\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)
Dấu "=" xảy ra khi \(\frac{a}{2}=b=c=d=e\)
4. Ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\)

\(\left(c-d\right)^2\ge0\Rightarrow c^2+d^2\ge2cd\)
\(\Rightarrow a^2+b^2+c^2+d^2\ge2ab+2cd\)

\(\Rightarrow a^2+b^2+c^2+d^2+ab+cd\ge3ab+3cd\)
Ta lại có:\(\left(\sqrt{ab}-\sqrt{cd}\right)^2\ge0\Rightarrow ab+cd\ge2\sqrt{abcd}=2\)

\(\Rightarrow3\left(ab+cd\right)\ge6\)
\(\Rightarrow a^2+b^2+c^2+d^2+ab+cd\ge3\left(ab+cd\right)\ge6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}a=b\\c=d\\ab=cd\end{cases}}\Leftrightarrow a=b=c=d\)

a ) \(x^2+4y^2+3z^2+14\ge2x+12y+6z\)

\(\Leftrightarrow x^2-2x+1+4y^2-12y+9+3z^2-6z+3+1\ge0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+3\left(z-1\right)^2+1\ge0\)

\(\LeftrightarrowĐPCM.\)

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

\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)

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

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

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

\(\LeftrightarrowĐPCM.\)

a) \(x^2+4y^2+3z^2+14\ge2x+12y+6z\)

\(\Rightarrow x^2+4y^2+3z^2+14-2x-12y-6z\ge0\)

\(\Rightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+3\left(z^2-2z+1\right)+1\ge0\)

\(\Rightarrow\left(x-1\right)^2+\left(2y-3\right)^2+3\left(z-1\right)^2\ge-1\)

Xem lại đề

b)

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

\(\Rightarrow3a^2+3b^2+3c^2\ge\left(a+b+c\right)^2\)

\(\Rightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) *Đúng*

Dấu "=" xảy ra khi: \(a=b=c\)