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điện thoại cùi nên chụp hơi mờ, đề này còn thiếu a,,bc>0
Lời giải:
Ta có:
\(\text{VT}=\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\)
\(=\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ac+a+b+c}{abc+(ab+bc+ac)+(a+b+c)+1}\)
\(=\frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\)
Ta cần chứng minh \(\text{VT}\geq \frac{3}{4}\)
\(\Leftrightarrow \frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\geq \frac{3}{4}\)
\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3(ab+bc+ac+a+b+c)+6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\)
\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\sqrt[6]{ab.bc.ac.a.b.c}\)
(Đúng theo BĐT Cô-si)
Do đó ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=1\)
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
Giải:
\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)
\(\Leftrightarrow\) \(\dfrac{a\left(c+1\right)+b\left(a+1\right)+c\left(b+1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)
Do a+1 ; b+1; c+1 >0
\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3
\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0
Áp dụng BĐT Cô-si cho 3 số
Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)
ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3
\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)
\(\Rightarrow\) (*) được chứng minh
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1
Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6
Ta có: bc(a2+1) = (a+b)(a+c)
\(\Rightarrow\) \(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}\) =\(\sqrt{\dfrac{a}{a+b}}.\sqrt{\dfrac{a}{a+c}}\)
Áp dụng BĐT Cô-si: \(\sqrt{\dfrac{a}{a+b}}.\sqrt{\dfrac{a}{a+c}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+c}+\dfrac{a}{a+c}\right)\)
\(\Rightarrow\) \(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
CMTT: \(\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{a+c}\right)\)
\(\dfrac{c}{\sqrt{ab\left(1+c^2\right)}}\) \(\le\) \(\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\) S \(\le\) \(\dfrac{1}{2}\left(\dfrac{a}{b+a}+\dfrac{a}{c+a}+\dfrac{b}{a+b}+\dfrac{b}{c+b}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)
\(\Rightarrow\) S\(\le\) \(\dfrac{1}{2}.3=\dfrac{3}{2}\)
Vậy Smax = \(\dfrac{3}{2}\)
Dấu "=" xảy ra\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=b=c\\a+b+c=abc\end{matrix}\right.\)
\(\Leftrightarrow\) \(a=b=c=\sqrt{3}\)
Áp dụng BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)
Ta có \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
Cộng VTV \(\Leftrightarrow3\ge\dfrac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\sqrt[3]{abc}+1\)
\(\Leftrightarrow VT^2=\sum\left[\dfrac{1}{a\left(b+1\right)}\right]^2\ge3\cdot\sum\dfrac{1}{ab\left(a+1\right)\left(b+1\right)}\\ \Leftrightarrow VT^2\ge3\cdot\dfrac{a^2+b^2+c^2+a+b+c}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\cdot\dfrac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\\ \Leftrightarrow VT^2\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(1+\sqrt[3]{abc}\right)^3}\\ \Leftrightarrow VT^2\ge\dfrac{9}{\sqrt[3]{\left(abc\right)^2}\left(1+\sqrt[3]{abc}\right)^2}=VP^2\\ \LeftrightarrowĐpcm\)
Dấu \("="\Leftrightarrow a=b=c=1\)