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Ta có BĐT : \(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{4}{a+b}\) ( \(a,b>0\) )
\(\dfrac{1}{b}+\dfrac{1}{c}\text{≥}\dfrac{4}{b+c}\left(b;c>0\right)\)
\(\dfrac{1}{a}+\dfrac{1}{c}\text{≥}\dfrac{4}{a+c}\left(a;c>0\right)\)
Cộng từng vế của các BĐT trên , ta có :
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{a+c}\)
⇔ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{≥}\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{a+c}\)
Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế theo vế ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
\(\Rightarrowđpcm\)
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
\(\dfrac{a^3}{1+b}+\dfrac{1+b}{4}+\dfrac{1}{2}\ge3\sqrt[3]{\dfrac{a^3\left(1+b\right)}{8\left(a+b\right)}}=\dfrac{3a}{2}\)
\(\dfrac{b^3}{1+c}+\dfrac{1+c}{4}+\dfrac{1}{2}\ge\dfrac{3b}{2}\) ; \(\dfrac{c^3}{1+a}+\dfrac{1+a}{4}+\dfrac{1}{2}\ge\dfrac{3c}{2}\)
\(\Rightarrow VT+\dfrac{a+b+c}{4}+\dfrac{9}{4}\ge\dfrac{3}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{5}{4}\left(a+b+c\right)-\dfrac{9}{4}\ge\dfrac{5}{4}.3\sqrt[3]{abc}-\dfrac{9}{4}=\dfrac{3}{2}\)
Lời giải:
Từ \(a+b+c\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow a+b+c\geq \frac{ab+bc+ac}{abc}\Rightarrow abc(a+b+c)\geq ab+bc+ac\)
\(\Rightarrow a^2b^2c^2(a+b+c)^2\geq (ab+bc+ac)^2(1)\)
Áp dụng BĐT AM-GM:
\(a^2b^2+b^2c^2\geq 2ab^2c\)
\(b^2c^2+c^2a^2\geq 2abc^2\)
\(a^2b^2+c^2a^2\geq 2a^2bc\)
Cộng theo vế, rút gọn \(\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)\)
\(\Rightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)(2)\)
Từ \((1);(2)\Rightarrow a^2b^2c^2(a+b+c)^2\geq 3abc(a+b+c)\)
\(\Rightarrow abc(a+b+c)\geq 3\Rightarrow a+b+c\geq \frac{3}{abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{\left ( \frac{a}{bc} \right )^2}{\frac{1}{c}}+\frac{\left ( \frac{b}{ca} \right )^2}{\frac{1}{a}}+\frac{\left ( \frac{c}{ab} \right )^2}{\frac{1}{b}}\geq \frac{\left ( \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
\(\Leftrightarrow \text{VT}\geq \frac{\left ( \frac{a^2+b^2+c^2}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
Theo hệ quả của BĐT AM-GM thì:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{\left ( \frac{ab+bc+ac}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz và AM-GM ta có:
\(\text{VT}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+(ab+bc+ac)+\frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{ac}\)
\(\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+(ab+bc+ac)+\frac{(a+b+c)^2}{ab+bc+ac}\)
\(\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{(ab+bc+ac).\frac{(a+b+c)^2}{ab+bc+ac}}\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2(a+b+c)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c+(a+b+c)\)
\(\geq 6\sqrt[6]{\frac{1}{a}.\frac{1}{b}.\frac{1}{c}.a.b.c}+(a+b+c)=6+a+b+c\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
\(\Leftrightarrow\left(1+abc\right)\left(\dfrac{1}{a\left(1+b\right)}+\dfrac{1}{b\left(1+c\right)}+\dfrac{1}{c\left(1+a\right)}\right)\ge3\)
Ta có:
\(\left(1+abc\right).\dfrac{1}{a\left(1+b\right)}=\dfrac{1+abc}{a+ab}=\dfrac{1+a+ab+abc-a-ab}{a+ab}=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}-1\)
\(\Rightarrow VT=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}+\dfrac{1+b}{b\left(1+c\right)}+\dfrac{c\left(1+a\right)}{1+c}+\dfrac{1+c}{c\left(1+a\right)}+\dfrac{a\left(1+b\right)}{1+a}-3\)
\(VT\ge6\sqrt[6]{\dfrac{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}-3=3\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
a) \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) + \(\dfrac{1}{c}\)≥\(\dfrac{9}{a+b+c}\)
<=> ( \(\dfrac{1}{a}\)+ \(\dfrac{1}{b}\) + \(\dfrac{1}{c}\))(a+b+c) ≥ 9
Ta có : \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) + \(\dfrac{1}{c}\) ≥ 3.căn bậc 3 1/abc(Cô-si)
a+b+c ≥ 3 căn bậc 3 abc
(1/a + 1/b + 1/c)(a+c+c) ≥ 9 căn bậc 3 abc/abc = 9
<=> 1/a + 1/b + 1/c ≥ 9(a+b+c)
Dấu ''='' xảy ra khi : a=b =c
Cách khác :
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\)
\("="\Leftrightarrow a=b=c\)