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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
=>\(\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)
=> (bc+ac+ab)(a+b+c)=abc
=> abc+b2c+bc2+a2c+abc+ac2+a2b+ab2+abc=abc
=>abc+b2c+bc2+a2c+abc+ac2+a2c+ab2+abc-abc=0
=>(a2c+2abc+b2c)+(a2b+ab2)+(ac2+bc2)=0
=>c(a+b)2+ab(a+b)+c2(a+b)=0
=>(a+b)[c(a+b)+ab+c2]=0
=>(a+b)(ac+bc+ab+c2)=0
=>(a+b)[a(c+b)+c(b+c)]=0
=>(a+b)(c+b)(a+c)=0
=> a+b=0, c+b=0, a+c=0
nếu a+b=0=>a=-b
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{-b^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{1}{c^3}\)(1)
và \(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\) (2)
từ (1) và (2) suy ra đpcm
Đặt vế trái BĐT cần chứng minh là P
Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ( Tự chứng minh BĐT này ), ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Rightarrow\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\le\dfrac{1}{\dfrac{4}{a+b}}=\dfrac{a+b}{4}\left(1\right)\)
Tương tự: \(\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}\le\dfrac{b+c}{4}\left(2\right)\)
\(\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\le\dfrac{c+a}{4}\left(3\right)\)
Cộng \(\left(1\right),\left(2\right),\left(3\right)\) vế theo vế, ta được:
\(P\le\dfrac{a+b+b+c+c+a}{4}=\dfrac{a+b+c}{2}\)
Dấu ''='' xảy ra khi và chỉ khi a=b=c
\(a+b+c=3abc\Rightarrow\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{ab}=3\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=9\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=9\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\cdot3=9\)
Vậy \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\).
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 \(\Leftrightarrow a=b=c=1\)
Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) và BĐT cần chứng minh là:
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel và AM-GM ta có:
\(VT=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}=VP\)
Xảy ra khi \(x=y=z=1 \Rightarrow a=b=c=1\)
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nhớ giải câu hỏi nhé ! thanks
Có gì đâu nhỉ?
Cauchy-Schwarz:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}=\dfrac{9}{2\left(a+b+c\right)}=\dfrac{4,5}{a+b+c}>\dfrac{3}{a+b+c}\)
áp dụng BĐT cauchy- schwarz ta có
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\)
⇔ \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2\left(a+b+c\right)}\)
⇔ \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{3}{a+b+c}\) (đpcm)
Ta chứng minh BĐT sau:
\(\dfrac{1}{x^3+x+2}\ge\dfrac{-x^2+3}{8}\) với \(x>0\)
Thật vậy, BĐT tương đương:
\(\left(x^2-3\right)\left(x^3+x+2\right)+8\ge0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^3+2x^2+x+2\right)\ge0\) (luôn đúng)
Áp dụng:
\(\Rightarrow VT\ge\dfrac{-a^2+3}{8}+\dfrac{-b^2+3}{8}+\dfrac{-c^2+3}{8}=\dfrac{9-\left(a^2+b^2+c^2\right)}{8}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)