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1. (a+b)^2 ≥ 4ab
<=> a2+2ab+b2≥ 4ab
<=> a2+2ab+b2-4ab≥ 0
<=> a2-2ab+b2≥ 0
<=> (a-b)^2 ≥ 0 ( luôn đúng )
2. a^2 + b^2 + c^2 ≥ ab + bc + ca
<=> 2a^2 + 2b^2 + 2c^2 ≥ 2ab + 2bc + 2ca
<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0
<=> (a^2- 2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2) ≥ 0
<=> (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0 ( luôn đúng)
Theo giả thiết, ta có: \(ab+bc+ca+abc=4\)
\(\Leftrightarrow abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8\)\(=12+\left(ab+bc+ca\right)+4\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\)\(=\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)\)
\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)
\(\Rightarrow a+b+c+6=12\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)-6+a+b+c\)
\(=\left(\frac{12}{a+2}+a-2\right)+\left(\frac{12}{b+2}+b-2\right)+\left(\frac{12}{c+2}+c-2\right)\)
Mặt khác: \(\frac{12}{a+2}+a-2=\frac{12+a^2-4}{a+2}=\frac{a^2+8}{a+2}\)
Tương tự: \(\frac{12}{b+2}+b-2=\frac{b^2+8}{b+2}\); \(\frac{12}{c+2}+c-2=\frac{c^2+8}{c+2}\)
Từ đó suy ra \(a+b+c+6=\frac{a^2+8}{a+2}+\frac{b^2+8}{b+2}+\frac{c^2+8}{c+2}\)
\(\ge\frac{\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2}{a+b+c+6}\)(Theo BĐT Bunyakovsky dạng phân thức)
\(\Rightarrow\left(a+b+c+6\right)^2\ge\left(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\right)^2\)
hay \(\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}\le a+b+c+6\)
Đẳng thức xảy ra khi a = b = c = 1
a) \(a^2+b^2=a^2+\frac{1}{4}+b^2+\frac{1}{4}-\frac{1}{2}\)
\(\ge2\sqrt{a^2.\frac{1}{4}}+2\sqrt{b^2.\frac{1}{4}}-\frac{1}{2}\) (bdt cosi)
\(=a+b-\frac{1}{2}=1-\frac{1}{2}=\frac{1}{2}\) (vi a+b=1)
dau = xay ra <=> a=b=1/2
chuc ban hoc tot
mik phai di ngu nen lam hoi tat mong bn thong cam
phan b bn lam tuong tu nha
1/ Ta có:
\(\left(a-b\right)^2\ge0,\) mọi a, b
<=> \(a^2-2ab+b^2\ge0\)
<=> \(2a^2+2b^2\ge a^2+2ab+b^2\)
<=> \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
<=> \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}=\frac{1}{2}\)
Dấu bằng xảy ra <=> a - b = 0 <=> a = b.
2/ Dựa vào câu 1.
\(a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(\frac{1}{2}\right)^2}{2}=\frac{1}{8}\).
a) Ta có: \(x^2-20x+101=x^2-2.x.10+10^2+1=\left(x-10\right)^2+1\)
Vì \(\left(x-10\right)^2\ge0\left(\forall x\in Z\right)\)
\(\Rightarrow\left(x-10\right)^2+1>1>0\)
Vậy x2-20x+101 >0 với mọi x
b) \(4a^2+4a+2=\left(2a\right)^2+2.2a.1+1+1=\left(2a+1\right)^2+1\)
Vì \(\left(2a+1\right)^2\ge0\left(\forall a\in Z\right)\)
\(\Rightarrow\left(2a+1\right)^2+1>1>0\)
Vậy 4a2+4a+2 > 0 với mọi a
c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+20\right)^2\) \(\ge0\left(\forall x\right)\)
a) \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab=\left(a+b\right)^2=2^2=4\)
\(\Leftrightarrow a^2+b^2\ge2\).
Dấu \(=\)khi \(a=b=1\).
b) \(\left(a^2-b^2\right)\ge0\Leftrightarrow a^4+b^4\ge2a^2b^2\Leftrightarrow2\left(a^4+b^4\right)\ge a^4+b^4+2a^2b^2=\left(a^2+b^2\right)^2\ge2^2=4\)
\(\Leftrightarrow a^4+b^4\ge2\)
Dấu \(=\)khi \(a=b=1\).
c) Bạn làm tương tự.
bạn j ơi a^2+b^2 có = 2 đâu