a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

c) \(C=4x-10-x^2=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2+6\right]\)

\(=-\left(x^2-4x+4+6\right)=-\left[\left(x-2\right)^2\right]-6\le-6< 0\forall x\)

\(B=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\)

\(A=x^2-xy+y^2\)

\(\Rightarrow A=x^2-xy+\dfrac{1}{4}y^2-\dfrac{1}{4}y^2+y^2\)

\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\)

mà \(\left(x-\dfrac{1}{2}y\right)^2\ge0;\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow A=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2\ge0\)

\(\Rightarrow\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\) với mọi x,y không đồng thời bằng 0

a)  \(A=x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)       với mọi x

b)   \(B=x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\) với mọi x

c)  \(x^2+xy+y^2+1=\left(x+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1>0\)  với mọi x,y

d)  bạn kiểm tra lại đề câu d) nhé:

 \(x^2+4y^2+z^2-2x-6y+8z+15\)

\(=\left(x-1\right)^2+\left(2y-\frac{6}{4}\right)^2+\left(z+4\right)^2-\frac{13}{4}\)

Đề câu d đúng mà!

\(A=x^2+x+1=x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

vậy A luôn luôn dương với mọi x

b: \(B=x^2-xy+y^2\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2\)

\(=\left(x-\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>0\forall x,y\ne0\)

c: \(C=-x^2+4x-10\)

\(=-\left(x^2-4x+10\right)\)

\(=-\left(x^2-4x+4+6\right)\)

\(=-\left(x-2\right)^2-6< 0\)

Lời giải:
Áp dụng BĐT AM-GM:

\(\frac{x^2}{y}+\frac{y^2}{x}+\sqrt{xy}=\frac{x^3+y^3}{2xy}+\frac{x^3+y^3}{2xy}+\sqrt{xy}\geq 3\sqrt[3]{\frac{(x^3+y^3)^2}{4xy\sqrt{xy}}}\)

Bằng BĐT AM-GM, dễ thấy:

\(x^3+y^3\geq \frac{1}{2}(x+y)(x^2+y^2)\geq \sqrt{xy}(x^2+y^2)\)

\(\Rightarrow (x^3+y^3)^2\geq xy(x^2+y^2)^2=xy\sqrt{x^2+y^2}.\sqrt{(x^2+y^2)^3}\geq xy\sqrt{2xy}\sqrt{(x^2+y^2)^3}\)

\(\Rightarrow \frac{x^2}{y}+\frac{y^2}{x}+\sqrt{xy}\geq 3\sqrt[3]{\frac{\sqrt{2}(x^2+y^2)^{\frac{3}{2}}}{4}}=3\sqrt{\frac{x^2+y^2}{2}}\)

Ta có đpcm

Dấu "=" xảy ra khi $x=y$

a) Ta có:

\(x^2-x+1\)

\(=x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Mà: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) và \(\dfrac{3}{4}>0\) nên

\(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

\(\Rightarrow x^2-x+1>0\forall x\)

a: Ta có: \(a^2+b^2+c^2=ab+bc+ac\)

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

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

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

=>a=b=c

b: ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)

Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)