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Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\c-a=z\end{matrix}\right.\Leftrightarrow x+y+z=0\)
\(\Leftrightarrow A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2\left(x+y+z\right)}{xyz}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-\dfrac{2\cdot0}{xyz}}\\ \Leftrightarrow A=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\left(đpcm\right)\)
BĐT đã cho tương đương với:
\(\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2-2\left[\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}\right]\ge2\left(\cdot\right)\).
Mặt khác ta có: \(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\).
Do đó \(\left(\cdot\right)\Leftrightarrow\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2\ge0\) (luôn đúng).
BĐT đã cho dc c/m.
Trước hết ta có:
\(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(b-c\right)\left(a-b\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{ab\left(a-b\right)+b^2c-a^2c+ac^2-bc^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
Do đó:
\(\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2-2+2\)
\(=\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2+2\left(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(a-b\right)\left(b-c\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\right)+2\)
\(=\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2+2\ge2\) (đpcm)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Nhìn qua đã biết là đề sai rồi bạn
Cho \(a,b,c\) các giá trị lớn ví dụ \(a=b=c=2\) là thấy sai ngay
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\dfrac{a^2}{2}+\dfrac{b^2}{c}+\dfrac{c^2}{c}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)
\(\Leftrightarrow a^2-\dfrac{a^2}{2}+b^2-\dfrac{b^2}{2}+c^2-\dfrac{c^2}{2}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{2\left(a^2+b^2+c^2+ab+bc+ca\right)}{4}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{4}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
Tương tự ta có \(\left\{{}\begin{matrix}\left(b+c\right)^2\ge4bc\\\left(c+a\right)^2\ge4ca\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2c+\left(a+b\right)^2\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2a+\left(b+c\right)^2\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2b+\left(c+a\right)^2\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2\left(c+1\right)\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2\left(a+1\right)\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2\left(b+1\right)\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}\le\dfrac{8}{4abc+\left(a+b\right)^2}\\\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}\le\dfrac{8}{4abc+\left(b+c\right)^2}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}\le\dfrac{8}{4abc+\left(c+a\right)^2}\end{matrix}\right.\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\) (3)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{\left(a+b\right)^2}{4}\ge2\sqrt{\dfrac{2}{c+1}}=\dfrac{4}{\sqrt{2\left(c+1\right)}}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{\left(b+c\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(a+1\right)}}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(c+a\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(b+1\right)}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\ge\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\)(4)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt{2\left(c+1\right)}\le\dfrac{c+3}{2}\)
\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}\ge\dfrac{8}{c+3}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2\left(a+1\right)}}\ge\dfrac{8}{a+3}\\\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{b+3}\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) (5)
Từ điều (3) , (4) , (5)
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2+4abc}+\dfrac{8}{\left(b+c\right)^2+4abc}+\dfrac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) ( đpcm )
\(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}=\dfrac{\left(\dfrac{a}{b+c}\right)^2}{a}+\dfrac{\left(\dfrac{b}{c+a}\right)^2}{b}+\dfrac{\left(\dfrac{c}{a+b}\right)^2}{c}\)
\(\ge\dfrac{\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2}{a+b+c}\ge\dfrac{\left(\dfrac{3}{2}\right)^2}{1}=\dfrac{9}{4}\)
Ta có:
\(BĐT\Leftrightarrow\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(a+c\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\dfrac{9}{4}\)
Rồi giải tương tự bài này:
Câu hỏi của phạm thảo - Toán lớp 10 | Học trực tuyến
Bài này trong đề nào đó mới đây:
Đặt \(\dfrac{a+b}{a-b}=x;\dfrac{b+c}{b-c}=y;\dfrac{c+a}{c-a}=z\).
Ta có: \(2P=\dfrac{\left(a-b\right)^2+\left(a+b\right)^2}{\left(a-b\right)^2}+\dfrac{\left(b-c\right)^2+\left(b+c\right)^2}{\left(b-c\right)^2}+\dfrac{\left(c-a\right)^2+\left(c+a\right)^2}{\left(c-a\right)^2}=3+x^2+y^2+z^2=3+\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\),
Mặt khác dễ dàng chứng minh được: \(\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(x-1\right)\left(y-1\right)\left(z-1\right)\Leftrightarrow xy+yz+zx=-1\).
Từ đó \(2P=\left(x+y+z\right)^2+5\ge5\Leftrightarrow P\ge\dfrac{5}{2}\).
Bài này là bất đẳng thức nên mình không tìm điểm rơi.