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Ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{1^2}{a^3\left(b+c\right)}+\dfrac{1^2}{b^3\left(c+a\right)}+\dfrac{1^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{a^2b^2c^2}{a^3\left(b+c\right)}+\dfrac{a^2b^2c^2}{b^3\left(c+a\right)}+\dfrac{a^2b^2c^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{b^2c^2}{a\left(c+b\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Svacxo ta có:
\(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)^2}{a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)}\) \(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)}{2}\) (1)
Chứng minh: \(\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\Leftrightarrow ab+bc+ca\ge3\)
Áp dụng BĐT Cosi ta có:
\(ab+bc+ca\ge3\sqrt[3]{ab.bc.ca}\)
\(ab+bc+ca\ge3\) (2)
Từ (1) và (2)
=> ĐPCM
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\Rightarrow ab+bc+ac=1\)
Ta có \(1+a^2=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự ta được \(1+b^2=\left(a+b\right)\left(b+c\right)\); \(1+c^2=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow A=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow A=\left[\left(a+b\right)\left(a+c\right)\left(b+c\right)\right]^2\) \(\Rightarrow A\) là số chính phương
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
\(a^3+b^3+c^3=3abc\\ \Rightarrow a^3+b^3+c^3-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\left(a+b+c\ne0\right)\\ \Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\\ \Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\\ \Rightarrow a=b=c\\ \Rightarrow B=\dfrac{2}{a}.\dfrac{2}{b}.\dfrac{2}{c}=\dfrac{8}{abc}\)
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 \(\Leftrightarrow a=b=c=1\)