1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
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 $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Bài 3:
Áp dụng BĐT AM-GM ta có:
\(\frac{a^2}{b^2}+1\geq 2\sqrt{\frac{a^2}{b^2}.1}=\frac{2a}{b}\). Tương tự:
\(\frac{b^2}{c^2}+1\geq \frac{2b}{c}; \frac{c^2}{a^2}+1\geq \frac{2c}{a}\)
Cộng các BĐT vừa thu được ta có :
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\geq 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
Mà theo BĐT AM-GM : \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}=3\)
Suy ra: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\geq 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\)
\(\Rightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
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Hoặc bạn có thể áp dụng BĐT Bunhiacopxky kết hợp AM-GM:
\(\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)(1+1+1)\geq \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq 3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Rightarrow \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\) (đpcm)
Bài 4:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)^2\)
Mà theo kết quả bài 3 thì ta có:
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\). Do đó:
\(\left(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq \left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Rightarrow \frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\geq \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
Bài 5:
Áp dụng BĐT AM-GM:
\((a+b+c)^2.\text{VT}=\frac{(a+b+c)^2}{a(a+b)}+\frac{(a+b+c)^2}{b(b+c)}+\frac{(a+b+c)^2}{c(c+a)}\)
\(=\frac{(a+b)^2+c^2+2c(a+b)}{a(a+b)}+\frac{a^2+(b+c)^2+2a(b+c)}{b(b+c)}+\frac{b^2+(c+a)^2+2b(c+a)}{c(c+a)}\)
\(=\frac{\frac{1}{4}(a+b)^2+c^2+2c(a+b)+\frac{3}{4}(a+b)^2}{a(a+b)}+\frac{a^2+\frac{1}{4}(b+c)^2+2a(b+c)+\frac{3}{4}(b+c)^2}{b(b+c)}+\frac{b^2+\frac{1}{4}(c+a)^2+2b(c+a)+\frac{3}{4}(c+a)^2}{c(c+a)}\)
\(\geq \frac{c(a+b)+2c(a+b)+\frac{3}{4}(a+b)^2}{a(a+b)}+\frac{a(b+c)+2a(b+c)+\frac{3}{4}(b+c)^2}{b(b+c)}+\frac{b(c+a)+2b(c+a)+\frac{3}{4}(c+a)^2}{c(c+a)}\)
\(=\frac{3c}{a}+\frac{3(a+b)}{4c}+\frac{3a}{b}+\frac{3(b+c)}{4b}+\frac{3b}{c}+\frac{3(c+a)}{4c}\)
\(=3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{3}{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\right)\)
\(\geq 3.3\sqrt[3]{1}+\frac{3}{4}.6\sqrt[6]{1}=\frac{27}{2}\)
Suy ra : \(\text{VT}\geq \frac{27}{2(a+b+c)^2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Cách khác bài 5:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}\geq 3\sqrt[3]{\frac{1}{a(a+b)}.\frac{1}{b(b+c)}.\frac{1}{c(c+a)}}=3\sqrt[3]{\frac{1}{a(b+c)b(c+a)c(a+b)}}=\frac{3}{\sqrt[3]{(ab+ac)(bc+ba)(ca+cb)}}(*)\)
Mà cũng theo BĐT AM-GM:
\((ab+ac)(bc+ba)(ca+cb)\leq \left(\frac{ab+ac+bc+ba+ca+cb}{3}\right)^3=\frac{8}{27}(ab+bc+ac)^3\)
\(ab+bc+ac\leq \frac{(a+b+c)^2}{3}\)
\(\Rightarrow (ab+ac)(bc+ba)(ca+cb)\leq \frac{8}{27}.\frac{(a+b+c)^6}{27}(**)\)
Từ \((*); (**)\Rightarrow \text{VT}\geq \frac{3}{\sqrt[3]{\frac{8}{27}.\frac{(a+b+c)^6}{27}}}=\frac{27}{2(a+b+c)^2}\)
Ta có đpcm.
