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Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1
Không mất tính tổng quát, giả sử đó là a và b
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)
\(\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)
\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)
Lại có:
\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)
\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)
\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
Lời giải:
Áp dụng BĐT AM-GM ta có:
$\text{VT}=[\frac{a+1}{4}+\frac{1}{a+1}+\frac{3}{4}a-\frac{1}{4}][\frac{b+1}{4}+\frac{1}{b+1}+\frac{3}{4}b-\frac{1}{4}][\frac{c+1}{4}+\frac{1}{c+1}+\frac{3}{4}c-\frac{1}{4}]$
$\geq [2\sqrt{\frac{1}{4}}+\frac{3}{4}a-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}b-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}c-\frac{1}{4}]$
$=\frac{3}{4}(a+1).\frac{3}{4}(b+1).\frac{3}{4}(c+1)$
$=\frac{27}{64}(a+1)(b+1)(c+1)$
$\geq \frac{27}{64}.2\sqrt{a}.2\sqrt{b}.2\sqrt{c}$
$=\frac{27}{64}.8\sqrt{abc}\geq \frac{27}{64}.8=\frac{27}{8}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a\left(b+c\right)}{4}\ge2\sqrt{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a\left(b+c\right)}{4}=2\sqrt{\dfrac{1}{4a^2}=\dfrac{1}{a}=\dfrac{abc}{a}=bc}}\)
Tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b\left(c+a\right)}{4}\ge\dfrac{1}{b}=ac\)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c\left(a+b\right)}{4}\ge\dfrac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow VT+\dfrac{ab+bc+ac}{2}\ge ab+bc+ac\)
\(\Rightarrow VT\ge\dfrac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\ge3^3\sqrt{a^2b^2c^2}=3\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu bằng xảy ra khi a=b=c=1
Bài này đã có ở đây:
Cho abc=1CMR\(\dfrac{a+3}{\left(a+1\right)^2}+\dfrac{b+3}{\left(b+1\right)^2}+\dfrac{c+3}{\left(c+1\right)^2}\ge3\) - Hoc24
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)
~*~*~*~*~
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\left(1-\dfrac{b}{b+c}\right)+\dfrac{b}{b+c}\left(1-\dfrac{c}{c+a}\right)+\dfrac{c}{a+c}\left(1-\dfrac{a}{a+b}\right)\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng
=> (1) đúng
Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).
Lời giải:
Đặt biểu thức vế trái là A
Có \(a+\frac{1}{a+1}=\frac{a^2+a+1}{a+1}=\frac{a^2}{a+1}+1=\frac{a^2}{a+1}+\frac{1}{2}+\frac{1}{2}\)
Áp dụng BĐT Cauchy-Schwarz:
\(a+\frac{1}{a+1}\geq \frac{(a+1+1)^2}{a+1+2+2}=\frac{(a+2)^2}{a+5}\)
Thực hiện tương tự với các phân thức còn lại và nhân theo vế:
\(\Rightarrow A\geq \frac{(a+2)^2(b+2)^2(c+2)^2}{(a+5)(b+5)(c+5)}\)
Áp dụng BĐT AM-GM:
\((a+2)(b+2)(c+2)\geq 3\sqrt[3]{a}.3\sqrt[3]{b}.3\sqrt[3]{c}=27\sqrt[3]{abc}\geq 27\)
\(\Rightarrow A\geq \frac{27(a+2)(b+2)(c+2)}{(a+5)(b+5)(c+5)}\) (1)
Ta sẽ cm
\(\frac{27(a+2)(b+2)(c+2)}{(a+5)(b+5)(c+5)}\geq \frac{27}{8}(*)\Leftrightarrow 8(a+2)(b+2)(c+2)\geq (a+5)(b+5)(c+5)\)
\(\Leftrightarrow 8[abc+8+2(ab+bc+ac)+4(a+b+c)]\geq abc+125+5(ab+bc+ac)+25(a+b+c)\)
\(\Leftrightarrow 7abc+11(ab+bc+ac)+7(a+b+c)\geq 61\)
BĐT trên luôn đúng theo AM_GM:
\(7abc+11(ab+bc+ac)+7(a+b+c)\geq 7abc+33\sqrt[3]{a^2b^2c^2}+21\sqrt[3]{abc}\geq 7+33+21=61\)
Do đó (*) đúng.
Từ \((1);(2)\Rightarrow A\geq \frac{27}{8}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)