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Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)c+ab\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(ab+bc+ca+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0
=> Hoặc a=-b hoặc b=-c hoặc c=-a
Ko mất tổng quát, g/s a=-b
a) Ta có: vì a=-b thay vào ta được:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)
=> đpcm
b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[a^2+2ab+b^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Ta có: \(N=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{a\cdot b\cdot c}=-1\)
Trường hợp 2: a=b=c
\(\Leftrightarrow N=\dfrac{b+b}{b}\cdot\dfrac{a+a}{a}\cdot\dfrac{c+c}{c}=2\cdot2\cdot2=8\)
1, Ta có a^3+b^3+c^3=3abc
-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2
-> (a+b)3 + c^3 - 3ab(a+b+c)=0
-> (a+b+c). ((a+b)^2-(a+b).c+c^2)-3ab(a+b+c)=0
-> (a+b+c)(a^2+2ab+b^2-ac-bc+c^2-3ab)=0
Th1: a+b+c=0
->P= a+b/2 . b+c/2 . c+a/2
= (-c)(-a)(-b)/2=-1
TH2 a^2+b^2+c^2-ab-bc-ca=0
->2a^2+2b^2+2c^2-2ab-abc-2ac=0
->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
-> (a-b)^2+(a-c)^2+(b-c)^2=0
Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0
Dấu = xảy ra (=)a-b=0
b-c=0
a-c=0
-> a=b=c
->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8
Ta có: \(\left\{{}\begin{matrix}abc=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3b^3c^3=1\\\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\end{matrix}\right.\)
\(a^3b^3+b^3c^3+c^3a^3=a^3b^3c^3\left(\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}\right)=1.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
\(\Rightarrow S=\left(a^3b^3+b^3c^3+c^3a^3\right)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2\)
Lại có:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{-3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{-3}{ab}\left(\dfrac{-1}{c}\right)=\dfrac{3}{abc}=3\)
\(\Rightarrow S=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2=3^2=9\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Đặt \(P=\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(abc\right)^2}{a^3\left(b+c\right)}+\dfrac{\left(abc\right)^2}{b^3\left(c+a\right)}+\dfrac{\left(abc\right)^2}{c^3\left(a+b\right)}\)
\(P=\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ca\right)^2}{b\left(c+a\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
\(P\ge\dfrac{\left(bc+ca+ab\right)^2}{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}\) (BĐT B.C.S)
\(=\dfrac{ab+bc+ca}{2}\) \(\ge\dfrac{3\sqrt[3]{abbcca}}{2}=\dfrac{3}{2}\) (do \(abc=1\)).
ĐTXR \(\Leftrightarrow a=b=c=1\)
Lời giải:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng theo vế:
\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)