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Ta có: \(a^2+b^2\ge2ab\) và \(b^2+1\ge2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2ab+2b+2}=\frac{1}{2\left(ab+b+1\right)}\left(1\right)\)
Tương tự ta có: \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\left(2\right);\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ac+a+1\right)}\left(3\right)\)
Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\) ta có:
\(VT\le\frac{1}{2\left(ab+b+1\right)}+\frac{1}{2\left(bc+c+1\right)}+\frac{1}{2\left(ac+a+1\right)}\)
\(=\frac{1}{2}\left(\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}\right)\left(abc=1\right)\)
\(=\frac{1}{2}\left(\frac{ac+a+1}{ac+a+1}\right)=\frac{1}{2}=VP\) (ĐPCM)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Câu hỏi của Tôi Là Ai - Toán lớp 8 - Học toán với OnlineMath
Câu hỏi của Tôi Là Ai - Toán lớp 8 - Học toán với OnlineMath
Ta có:
\(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2ab+2b+2}=\frac{1}{2}\cdot\frac{1}{ab+b+1}\)
Tương tự CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1}\) và \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)=\frac{1}{2}\cdot1=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
A=\(\frac{1}{a^2+2b^2+3}\)+\(\frac{1}{b^2+2c^2+3}\)+\(\frac{1}{c^2+2a^2+3}\)
ta có: \(\frac{1}{a^2+2b^2+3}\)=\(\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\)\(\le\)\(\frac{1}{2\left(ab+b+1\right)}\)
vì : a2+b2\(\ge\)2\(\sqrt{a^2b^2}\)=2ab
b2+1\(\ge\)2\(\sqrt{b^2x1}\)=2b
cmtt => A\(\le\)\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{1}{bc+c+1}\)+\(\frac{1}{ca+a+1}\))
=\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab^2c+abc+ab}\)+\(\frac{b}{cba+ab+b}\))
=\(\frac{1}{2}\)x (\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab+b+1}\)+\(\frac{b}{ab+b+1}\))=\(\frac{1}{2}\)x\(\frac{ab+b+1}{ab+b+1}\)=\(\frac{1}{2}\)
dấu "=" xảy ra <=> a=b=c=1
Áp dụng BĐT Cô-si ta có:
\(a^2+b^2\ge2ab;b^2+1^2\ge2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2\)
\(\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}=\frac{1}{2}.\frac{1}{ab+b+1}\)
chứng minh tương tự
\(\Rightarrow\frac{1}{b^2+2c^2+3}\le\frac{1}{2}.\frac{1}{bc+c+1};\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)
đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}=\frac{ac+a+1}{ac+a+1}=1\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.1=2\)
=>đpcm
\(a\in\left[-2;5\right]\Rightarrow\left(a+2\right)\left(a-5\right)\le0\)
\(\Leftrightarrow a^2\le3a+10\)
Tương tự: \(b^2\le3b+10\Rightarrow2b^2\le6b+20\)
\(c^2\le3c+10\Rightarrow3c^2\le9c+30\)
Cộng vế:
\(a^2+2b^2+3c^2\le3\left(a+2b+3c\right)+60\le66\) (đpcm)