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Ta có \(a+b+c=abc\Leftrightarrow\dfrac{a+b+c}{abc}=1\) \(\Leftrightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Lại có \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\Leftrightarrow2^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\) (đpcm)
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
Em kiểm tra lại mẫu số của biểu thức c, chắc chắn đề sai
Cho a,b,c>2 thỏa mãn \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\).Chứng minh rằng:(a-2)(b-2)(c-2)≤1.
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
\(\Rightarrow\dfrac{1}{a}=\left(\dfrac{1}{2}-\dfrac{1}{b}\right)+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\)
Dễ dàng chứng minh \(\dfrac{b-2}{2b},\dfrac{c-2}{2c}\) là các số dương.
Áp dụng BĐT Cauchy cho 2 số dương ta có:
\(\dfrac{b-2}{2b}+\dfrac{c-2}{2c}\ge2\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{4bc}}=\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\)
\(\Rightarrow\dfrac{1}{a}\ge\sqrt{\dfrac{\left(b-2\right)\left(c-2\right)}{bc}}\left(1\right)\)
CMTT ta có: \(\left\{{}\begin{matrix}\dfrac{1}{b}\ge\sqrt{\dfrac{\left(c-2\right)\left(a-2\right)}{ca}}\left(2\right)\\\dfrac{1}{c}\ge\sqrt{\dfrac{\left(a-2\right)\left(b-2\right)}{ab}}\left(3\right)\end{matrix}\right.\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\dfrac{1}{abc}\ge\dfrac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{abc}\)
\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le1\left(đpcm\right)\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=c\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{matrix}\right.\Leftrightarrow a=b=c=3\)
a.
Bình phương 2 vế, BĐT cần chứng minh trở thành:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)
Ta có:
\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cộng lại:
\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
b.
\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)
Nên ta chỉ cần chứng minh:
\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)
\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)
Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)
Lời giải:
\(abc=(1-a)(1-b)(1-c)\Rightarrow \frac{1-a}{a}.\frac{1-b}{b}.\frac{1-c}{c}=1\)
Đặt \(\left(\frac{1-a}{a};\frac{1-b}{b}; \frac{1-c}{c}\right)=(x,y,z)\Rightarrow (a,b,c)=\left(\frac{1}{x+1}; \frac{1}{y+1}; \frac{1}{z+1}\right)\)
Bài toán trở thành
Cho $x,y,z>0$ thỏa mãn $xyz=1$. CMR:
\(A=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2}+\frac{1}{(z+1)^2}\geq \frac{3}{4}\)
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Thật vậy:
Áp dụng BĐT Bunhiacopxky:
\((x+1)^2\leq (x+\frac{1}{y})(x+y)\Rightarrow \frac{1}{(x+1)^2}\geq \frac{y}{(xy+1)(x+y)}\)
\((y+1)^2\leq (y+\frac{1}{x})(y+x)\Rightarrow \frac{1}{(y+1)^2}\geq \frac{x}{(xy+1)(x+y)}\)
\(\Rightarrow A\geq \frac{y}{(xy+1)(x+y)}+\frac{x}{(xy+1)(x+y)}+\frac{1}{(z+1)^2}\)
\(A\geq \frac{x+y}{(xy+1)(x+y)}+\frac{1}{(z+1)^2}=\frac{1}{xy+1}+\frac{1}{(z+1)^2}\)
\(A\geq \frac{1}{\frac{1}{z}+1}+\frac{1}{(z+1)^2}=\frac{z^2+z+1}{(z+1)^2}(*)\)
Mà \(\frac{z^2+z+1}{(z+1)^2}-\frac{3}{4}=\frac{(z-1)^2}{4(z+1)^2}\geq 0\Rightarrow \frac{z^2+z+1}{(z+1)^2}\geq \frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow A\geq \frac{3}{4}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c=\frac{1}{2}\)