\(x^2-\sqrt{x}+\dfrac{1}{2}\)

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

\(=\left(x-\dfrac{1}{2}\right)^2+\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\\sqrt{x}-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\) vô nghiệm

Vậy \(x^2-\sqrt{x}+\dfrac{1}{2}>0\forall x\ge0\)

a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)

a, Với x ≥ 0, x ≠1 
P= [ \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] : \(\dfrac{\sqrt{x}-1}{2}\)  = 
\(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)]
: \(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\):\(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
P= \(\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}\right)^2+2.\dfrac{1}{2}.\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\)= (\(\sqrt{x}+\dfrac{1}{2}\))² +\(\dfrac{3}{4}\) >\(0\)  ∀ x
=> \(\dfrac{3}{x+\sqrt{x}+1}>0\) ∀ x

=> P > 0 với mọi x ≥ 0 và x ≠ 1

Áp dụng bất đẳng thức Mincopski

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

Chứng minh rằng : \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)^2+9\ge6\left(x+y+z\right)\)

\(\Leftrightarrow\frac{\left(x+y+z\right)^2+9}{x+y+z}\ge6\)

\(\Leftrightarrow x+y+z+\frac{9}{x+y+z}\ge6\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow x+y+z+\frac{9}{x+y+z}\ge2\sqrt{\frac{9\left(x+y+z\right)}{x+y+z}}=2\sqrt{9}=6\left(đpcm\right)\)

Vậy \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

Mà \(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(x=y=z=1\)

Chúc bạn học tốt !!!

Áp dụng bất đẳng thức Mincopski

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

Chứng minh rằng : \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)^2+9\ge6\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z+\frac{9}{x+y+z}\ge6\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow x+y+z+\frac{9}{x+y+z}\ge2\sqrt{\frac{9\left(x+y+z\right)}{x+y+z}}=2\sqrt{9}=6\left(đpcm\right)\)

Vậy \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

Mà \(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(x=y=z=1\)

Chúc bạn học tốt !!!

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Áp dụng bất đẳng thức Mincopski

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

Chứng minh rằng \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

\(\Leftrightarrow\left(x+y+z\right)^2+9\ge6\left(x+y+z\right)\)

\(\Leftrightarrow\dfrac{\left(x+y+z\right)^2+9}{x+y+z}\ge6\)

\(\Leftrightarrow x+y+z+\dfrac{9}{x+y+z}\ge6\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow x+y+z+\dfrac{9}{x+y+z}\ge2\sqrt{\dfrac{9\left(x+y+z\right)}{x+y+z}}=2\sqrt{9}=6\) ( đpcm )

Vậy \(\sqrt{\left(x+y+z\right)^2+9}\ge\sqrt{6\left(x+y+z\right)}\)

Mà \(\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{\left(x+y+z\right)^2+9}\)

\(\Rightarrow\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\ge\sqrt{6\left(x+y+z\right)}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=1\)

a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)

b) \(x-x^2-3=-\left(x^2-x+3\right)\)

\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)

\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)

\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)

x²-2x+2=(x²-2x+1)+1=( x-1)²+1

Mà (x-1)²≥0 với mọi x

=> (x-1)²+1>0 với mọi x

=> x²-2x+2>0 với mọi x

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

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

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Rightarrow\sqrt{2\left(a^2+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=\left|a+b\right|\)

Dấu "=" \(\Leftrightarrow a=b\)

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x^2 + 2x + 2 = x^2 + 2.x.1 + 1^2 +1 = (x + 1)^2 + 1 > 0

-x^2 + 4x - 4 = -(x^2 - 2.x.2 + 2^2) = -(x - 2)^2 <= 0

a) ta co ; x^2+ 2x+ 2= (x²+2x+1)+1=(x+1)²+1>0

vi (x+1)²>hoặc=0;1>0suy ra x^2+ 2x+ 2>0

b)ta co  -x²+4x-4=-(x²-4x+4)=-(x-2)²<0

a) x^2 + 2x + 2 = ( x^2 + 2x +1 ) + 1 =( x + 1)^2 +1 >0 với mọi x 

b) -x^2 + 4x - 4 = -( x^2 -4x + 4 ) = - ( x - 2)^2  ≤ 0 với mọi x

a. \(B=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\\ =\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{2}\\ =\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

b.Ta có:

\(B=\dfrac{2}{x+\sqrt{x}+1}\). Mà \(\left[{}\begin{matrix}2>0\\x+\sqrt{x}+1=\left[\left(\sqrt{x}\right)^2+2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}\right]+\dfrac{3}{4}>0\end{matrix}\right.\)

Vậy B>0 \(\forall x\)