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Câu hỏi của tran duc huy - Toán lớp 10 | Học trực tuyến
\(a-b+b+\frac{1}{b\left(a-b\right)}\ge3\sqrt[3]{\frac{\left(a-b\right)b.1}{b\left(a-b\right)}}=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(VT=a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+\frac{b+1}{2}+\frac{b+1}{2}-1\)
\(VT\ge4\sqrt[4]{\frac{4\left(a-b\right)\left(b+1\right)^2}{4\left(a-b\right)\left(b+1\right)^2}}-1=3\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}b=1\\a=2\end{matrix}\right.\)
\(\frac{a-b}{2}+\frac{a-b}{2}+\frac{1}{b\left(a-b\right)^2}+b\ge4\sqrt[4]{\frac{b\left(a-b\right)^2}{4b\left(a-b\right)^2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=\frac{3\sqrt{2}}{2}\\b=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(BDT\Leftrightarrow\sum\left[\dfrac{\left(a+b\right)^2}{c^2+ab}-2\right]\ge0\)\(\Leftrightarrow\sum\dfrac{a^2+b^2-2c^2}{c^2+ab}\ge0\)(*)
\(\Leftrightarrow\sum\left(\dfrac{a^2-c^2}{c^2+ab}+\dfrac{b^2-c^2}{c^2+ab}\right)\ge0\)
\(\Leftrightarrow\sum\left(c^2-a^2\right)\left(\dfrac{1}{a^2+bc}-\dfrac{1}{c^2+ab}\right)\ge0\)
\(\Leftrightarrow\sum\left(c-a\right)^2.\dfrac{\left(c+a\right)\left(c+a-b\right)}{\left(a^2+bc\right)\left(c^2+ab\right)}\ge0\)
\(\dfrac{\left(a+b\right)^2}{c^2+ab}+\dfrac{\left(b+c\right)^2}{a^2+bc}+\dfrac{\left(c+a\right)^2}{b^2+ca}\ge\dfrac{\left(a+b+b+c+c+a\right)^2}{a^2+b^2+c^2+ab+bc+ca}\)\(=\dfrac{4\left(a+b+c\right)^2}{a^2+b^2+c^2+ab+bc+ca}\) (theo AM-GM với a ; b>0)
\(=\dfrac{4\left(a^2+b^2+c^2+2ab+2bc+2ca\right)}{a^2+b^2+c^2+ab+bc+ca}=\dfrac{4.3.\left(a^2+b^2+c^2\right)}{2.\left(a^2+b^2+c^2\right)}\)(do \(a^2+b^2+c^2\ge ab+bc+ca\))
\(=4.1,5\) = 6 ( do a;b;c>0)
\(a+\frac{4}{b\left(a-b\right)^2}=a-b+b+\frac{4}{b\left(a-b\right)^2}\ge a-b+2\sqrt{\frac{4b}{b\left(a-b\right)^2}}=a-b+\frac{4}{a-b}\ge4\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=3\\b=1\end{matrix}\right.\)
b/ \(a-b+\frac{4}{\left(a-b\right)\left(b+1\right)^2}+b\ge2\sqrt{\frac{4\left(a-b\right)}{\left(a-b\right)\left(b+1\right)^2}}+b=\frac{4}{b+1}+b+1-1\ge4-1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
am-gm là ra thoi bạn :v