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Ta có
a2 + b2 + c2 \(\ge\)ab + bc + ca
<=> 2(a2 + b2 + c2)\(\ge\)2(ab + bc + ac)
<=> 3(a2 + b2 + c2)\(\ge\)(a + b + c)2
<=> a2 + b2 + c2 \(\ge\frac{\left(a+b+c\right)^2}{3}\)= \(\frac{9}{4×3}=\frac{3}{4}\)
Đạt GTNN khi a = b = c = \(\frac{1}{2}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{bc+ac+ab}{abc}=0\Rightarrow bc+ac+ab=0\)
Biến đổi vế phải ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(=a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2+2.0=a^2+b^2+c^2\)
=> ĐPCM
B, -x^2 + 2x - 4 = - ( x^2 - 2x + 4 ) = - ( x^2 - 2x + 1 + 3 ) = -(x + 1 )^2 - 3 <= -3
=> 3/ -(x+1)^2-3 >= 3/-3=-1
Vậy GTNN của A là -1 khi x = -1
Ta có: \(abc\le\frac{\left(a+b+c\right)^3}{27}\) ; \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)
Mà \(a^2+b^2+c^2=3abc\)
=>\(\frac{\left(a+b+c\right)^2}{3}\le\frac{\left(a+b+c\right)^3}{27}.3\)
=> \(a+b+c\ge3\)
Áp dụng bđt bunhia dạng phân thức ta có:
\(M\ge\frac{\left(a+b+c\right)^2}{a+b+c+6}\)
Đặt \(a+b+c=x\left(x\ge3\right)\)
=> \(M\ge\frac{x^2}{x+6}\)
Xét \(\frac{x^2}{x+6}\ge\frac{5}{9}x-\frac{2}{3}\)
<=>\(x^2\ge\frac{5}{9}x^2+\frac{8}{3}x-4\)
<=>\(\left(\frac{2}{3}x-2\right)^2\ge0\)(luôn đúng)
=> \(M\ge\frac{5}{9}x-\frac{2}{3}\ge\frac{5}{9}.3-\frac{2}{3}=1\)
=>\(MinM=1\)xảy ra khi a=b=c=1
Có: \(\frac{ab}{c}\)+\(\frac{bc}{a}\)>= 2 .\(\left(\frac{ab.bc}{ac}\right)\)= 2b^2
Tương tự, => 2.(ab/c+bc/a+ac/b) >=2(a^2 + b^2 + c^2)
<=> ab/c+bc/a+ac/b >=1
Dấu "=" xảy ra <=> a=b=c=\(\frac{\sqrt{3}}{3}\)
a)\(A=3\cdot\left|1-2x\right|-5\)
Vì \(\left|1-2x\right|\ge0\Rightarrow3\cdot\left|1-2x\right|\ge0\Rightarrow3\cdot\left|1-2x\right|-5\ge0-5=-5\)
\(\Rightarrow A\ge-5\)
\(\Rightarrow MIN_A=-5\Leftrightarrow\left|1-2x\right|=0\Leftrightarrow1-2x=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b)\(B=\left(2x^2+1\right)^4-3\)
Vì \(\left(2x^2+1\right)^4\ge1\Rightarrow\left(2x^2+1\right)^4-3\ge1-3=-2\)
\(\Rightarrow A\ge-2\)
\(\Rightarrow MIN_A=-2\Leftrightarrow\left(2x^2+1\right)^4=1\Leftrightarrow2x^2+1=1\Leftrightarrow2x^2=0\Leftrightarrow x=0\)
c)\(C=\left|x-\frac{1}{2}\right|+\left(y+2\right)^2+11\)
Vì \(\left|x-\frac{1}{2}\right|\ge0,\left(y+2\right)^2\ge0\Rightarrow\left|x-\frac{1}{2}\right|+\left(y+2\right)^2+11\ge0+0+11=11\)
\(\Rightarrow A\ge11\)
\(\Rightarrow MIN_A=11\Leftrightarrow\left|x-\frac{1}{2}\right|=0\Leftrightarrow x=\frac{1}{2},\left(y+2\right)^2=0\Leftrightarrow y+2=0\Leftrightarrow y=-2\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+ac+bc\right)=\frac{9}{4}\)\(\Rightarrow2\left(ab+ac+bc\right)=\frac{9}{4}-\left(a^2+b^2+c^2\right)\)
mà ta có \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+ac+bc\right)\ge0\)\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-\frac{9}{4}+\left(a^2+b^2+c^2\right)\ge0\)
\(3\left(a^2+b^2+c^2\right)\ge\frac{9}{4}\Leftrightarrow\left(a^2+b^2+c^2\right)\ge\frac{3}{4}\)có \(\left(a^2+b^2+c^2\right)\)đạt min là 3/4 khi và chỉ khi a=b=c=1/2