Cho \(a^2+b^2+c^2=1\) va \(a^3+b^3+c^3=1\)
Tính \(a+b^2+c^3=?\)
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Sử dụng:
\(A^3+B^3+C^3-3ABC=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-AC\right)\) (1)
Áp dụng vào bài:
\(\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3-3\left(a-1\right)\left(b-2\right)\left(c-3\right)\)
\(=\left(a-1+b-2+c-3\right)\)[ \(\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2\)
\(+\left(a-1\right)\left(b-2\right)+\left(a-1\right)\left(c-3\right)+\left(b-2\right)\left(c-3\right)\)]
<=> \(0-3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)
( vì \(a-1+b-2+c-3=a+b+c-6=6-6=0\))
<=> \(\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)
<=> a = 1 hoặc b = 2 hoặc c = 3.
Không mất tính tổng quát: g/s : a = 1
Khi đó: b + c =5
Ta có: \(T=\left(b-2\right)^{2n+1}+\left(c-3\right)^{2n+1}\)
\(=\left(b-2+c-3\right).A\)
\(=\left(b+c-5\right).A\)
\(=0.A=0\)
Với \(A=\left(b-2\right)^{2n}-\left(b-2\right)^{2n-1}\left(c-3\right)+\left(b-2\right)^{2n-2}\left(c-3\right)^2-...+\left(c-3\right)^{2n}\)
Tương tự b = 2; c= 3 thì T = 0.
Vậy T = 0.
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Rightarrow2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Rightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(\frac{-1}{c}\right)^3\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}.\left(-\frac{1}{c}\right)=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{ab}=0\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\left(đpcm\right)\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\Rightarrow ab+bc+ac=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}=0\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(\frac{1}{a}\right)^3+\left(\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}\)
\(\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Ta có 1 + ab2 \(\ge\)\(2b\sqrt{a}\)
1 + bc2 \(\ge2c\sqrt{b}\)
1 + ca2 \(\ge2a\sqrt{c}\)
VT \(\ge\)\(2\left(\frac{b\sqrt{a}}{c^3}+\frac{c\sqrt{b}}{a^3}+\frac{a\sqrt{c}}{b^3}\right)\)
\(\ge2\frac{\left(\sqrt[4]{b^2a}+\sqrt[4]{c^2b}+\sqrt[4]{a^2c}\right)^2}{a^3+b^3+c^3}\)
\(\ge2\frac{\left(3\sqrt[12]{a^3b^3c^3}\right)^2}{a^3+b^3+c^3}\)
\(\ge\frac{18}{a^3+b^3+c^3}\)