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a) Áp dụng BĐT Cauchy-Schwarz ta có:
(12+12+12)(x2+y2+z2)≥(x+y+z)2(12+12+12)(x2+y2+z2)≥(x+y+z)2
⇒3(x2+y2+z2)≥(x+y+z)2⇒3(x2+y2+z2)≥(x+y+z)2
⇒3(x2+y2+z2)≥(x+y+z)2=12=1⇒3(x2+y2+z2)≥(x+y+z)2=12=1
⇒x2+y2+z2≥13⇒x2+y2+z2≥13
Đẳng thức xảy ra khi x=y=z=13x=y=z=13
b) Áp dụng BĐT Cauchy-Schwarz ta có:
(4+1)(4x2+y2)≥(4x+y)2(4+1)(4x2+y2)≥(4x+y)2
⇒5(4x2+y2)≥(4x+y)2⇒5(4x2+y2)≥(4x+y)2
⇒5(4x2+y2)≥(4x+y)2=12=1⇒5(4x2+y2)≥(4x+y)2=12=1
⇒4x2+y2≥15⇒4x2+y2≥15
Đẳng thức xảy ra khi x=y=15x=y=15
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\(x^2>=\dfrac{1}{4}\)
\(y^2>=\dfrac{1}{4}\)
Do đó: \(x^2+y^2>=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)
\(x\ge\dfrac{1}{2};y\ge\dfrac{1}{2}\)=>\(xy\ge\dfrac{1}{4}\)=>\(2xy\ge\dfrac{1}{2}\).
\(x+y\ge\dfrac{1}{2}+\dfrac{1}{2}=1\)
=>\(\left(x+y\right)^2\ge1\)
=>\(x^2+2xy+y^2\ge1\)
=>\(x^2+y^2\ge1-2xy\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)
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Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{2005^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2004.2005}\)
Mà \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2004.2005}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2004}-\frac{1}{2005}\)
\(=1-\frac{1}{2005}=\frac{2004}{2005}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{2005^2}< \frac{2004}{2005}\left(\text{đ}pcm\right)\)
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\(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\frac{k+2-k}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\left(\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)\left(k+2\right)}\right)\)
\(=\frac{1}{2}\left[\frac{k+1-k}{k\left(k+1\right)}-\frac{\left(k+2\right)-\left(k+1\right)}{\left(k+1\right)\left(k+2\right)}\right]\)
\(=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{k+1}+\frac{1}{k+2}\right)\)
\(=\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+2}\right)-\frac{1}{k+1}\)
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Cách 1:Ta có: \(2\left(1+a^2\right)\ge\left(1+a\right)^2\)
\(\Rightarrow\frac{1}{\left(1+a\right)^2}\ge\frac{1}{\left[2\left(1+a^2\right)\right]}\)
\(\Rightarrow\frac{1}{\left(1+x\right)^2}+\frac{1}{1+y^2}\ge\frac{1}{\left[2\left(1+x^2\right)\right]}+\frac{1}{\left[2\left(1+y^2\right)\right]}\)
mà: \(\frac{1}{1+x^2}+\frac{1}{1+y^2}=\frac{2+x^2+y^2}{1+x^2y^2+x^2+y^2}\)
\(\Rightarrow\frac{1}{1+x^2}+\frac{1}{1+y^2}=\frac{\left[2.\left(1+xy\right)+\left(x-y\right)^2\right]}{\left(1+xy\right)^2+\left(x-y\right)^2}\)
\(\Rightarrow\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge2.\frac{1+xy}{\left(1+xy\right)^2}\)
\(\Rightarrow\frac{1}{\left[2\left(1+x^2\right)\right]}+\frac{1}{\left[2\left(1+y^2\right)\right]}\ge\frac{1}{1+xy}\)
\(\Rightarrow\frac{1}{\left(1+x\right)^2}+\frac{1}{1+y^2}\ge\frac{1}{1+xy}\)
TL :
Vì 1 + 1 = 2
HT
.