Cho a,b,c > 0. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\)
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Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
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\(\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)
\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)
\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)
\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)
\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow\) đpcm
Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ
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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 $a=b=c$
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Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a+b}{ab+c^2}=\dfrac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\le\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\)
Tương tự:
\(\dfrac{b+c}{bc+a^2}\le\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}\) ; \(\dfrac{c+a}{ca+b^2}\le\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)
Cộng vế:
\(VT\le\dfrac{1}{a}\left(\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)+\dfrac{1}{b}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)+\dfrac{1}{c}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
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Lời giải:
Áp dụng BĐT Cô-si cho các số dương ta có:
\(\frac{a}{bc}+\frac{b}{ac}\geq 2\sqrt{\frac{a}{bc}.\frac{b}{ac}}=2\sqrt{\frac{1}{c^2}}=\frac{2}{c}\)
\(\frac{b}{ac}+\frac{c}{ab}\geq 2\sqrt{\frac{b}{ac}.\frac{c}{ab}}=2\sqrt{\frac{1}{a^2}}=\frac{2}{a}\)
\(\frac{a}{bc}+\frac{c}{ab}\ge 2\sqrt{\frac{a}{bc}.\frac{c}{ab}}=2\sqrt{\frac{1}{b^2}}=\frac{2}{b}\)
Cộng các BĐT trên theo vế và rút gọn
\(\Rightarrow \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
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Đầu tiên ta cm bđt:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)(tự cm)
Áp dụng ta có:
\(A=\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{9}{3+ab+bc+ca}\)
Cần cm:\(ab+bc+ca\le3\)
Hay \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
=>đpcm
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