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theo de bai ta co \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\) suy ra ab+bc+ac=abc
\(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ac}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
nên vt =\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+c\right)\left(c+b\right)}\)
nx \(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\) >= \(\dfrac{3a}{4}\)
ttu vt>= \(\dfrac{3\left(a+b+c\right)}{4}-\left(\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{a+b}{8}+\dfrac{b+c}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\right)\) =\(\dfrac{a+b+c}{4}\)
dau = say ra a=b=c=3
Đang rảnh, làm luôn\(A=\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}=\dfrac{1}{2}\left[\left(\dfrac{a}{bc}+\dfrac{b}{ca}\right)+\left(\dfrac{b}{ca}+\dfrac{c}{ab}\right)+\left(\dfrac{c}{ab}+\dfrac{a}{bc}\right)\right]\ge\dfrac{1}{2}\left(\dfrac{2}{c}+\dfrac{2}{a}+\dfrac{2}{b}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 2
Lời giải:
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ac=0\)
Khi đó:
\((\sqrt{a+c}+\sqrt{b+c})^2=a+c+b+c+2\sqrt{(a+c)(b+c)}\)
\(=a+b+2c+2\sqrt{ab+ac+bc+c^2}=a+b+2c+2\sqrt{c^2}\)
\(=a+b+2c+2|c|\)
Vì $a,b$ dương nên \(\frac{-1}{c}=\frac{1}{a}+\frac{1}{b}>0\Rightarrow c< 0\Rightarrow 2|c|=-2c\)
Do đó:
\((\sqrt{a+c}+\sqrt{b+c})^2=a+b+2c+2|c|=a+b+2c+(-2c)=a+b\)
\(\Rightarrow \sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\)
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Cách khác:
Ta chứng minh: \(\frac{a}{a^3+a+1}\le\frac{1}{2}.\frac{a^{\frac{2}{3}}+1}{a^{\frac{4}{3}}+a^{\frac{2}{3}}+1}\) (1)
Đặt \(a=x^3\Leftrightarrow\frac{x^3}{x^9+x^3+1}\le\frac{1}{2}.\frac{x^2+1}{x^4+x^2+1}\)
Tương đương với $$\frac{(x - 1)^2 (x^9 + 2 x^8 + 4 x^7 + 6 x^6 + 6 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)}{2 (x^2 - x + 1) (x^2 + x + 1) (x^9 + x^3 + 1)} \geq 0$$
Vậy (1) đúng. Thiết lập $3$ bất đẳng thức tương tự và cộng theo vế thu đượcVasc.
\(\Rightarrow\) $\text{đpcm}$
Đẳng thức xảy ra khi $a=b=c=1$
Lời giải:
Xét hiệu: $a^3+1-a(a+1)=a^2(a-1)-(a-1)=(a+1)(a-1)^2\geq 0$ với mọi $a>0$
$\Rightarrow a^3+1\geq a(a+1)\Rightarrow a^3+a+1\geq a(a+2)$
$\Rightarrow \frac{a}{a^3+a+1}\leq \frac{1}{a+2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế thu được:
$\sum \frac{a}{a^3+a+1}\leq \sum \frac{1}{a+2}(*)$
Do $abc=1$ nên tồn tại $x,y,z>0$ sao cho $(a,b,c)=(\frac{x^2}{yz}, \frac{y^2}{xz}, \frac{z^2}{xy})$
Khi đó, áp dụng BĐT Cauchy-Schwarz:
$\sum \frac{1}{a+2}=\sum \frac{yz}{x^2+2yz}=\frac{1}{2}\sum (1-\frac{x^2}{x^2+2yz})=\frac{3}{2}-\frac{1}{2}.\sum \frac{x^2}{x^2+2yz}\leq \frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{x^2+2yz+y^2+2xz+z^2+2xy}$
$=\frac{3}{2}-\frac{1}{2}.\frac{(x+y+z)^2}{(x+y+z)^2}=1(**)$
Từ $(*); (**)$ ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=1$
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
Ta có:\(\dfrac{1}{a+b+1}+\dfrac{1}{b+c+1}+\dfrac{1}{a+c+1}=2\)
\(\Rightarrow\dfrac{1}{a+b+1}=\left(1-\dfrac{1}{b+c+1}\right)+\left(1-\dfrac{1}{a+c+1}\right)\)
\(\Rightarrow\dfrac{1}{a+b+1}=\dfrac{b+c}{b+c+1}+\dfrac{a+c}{a+c+1}\ge2\sqrt{\dfrac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)Chứng minh tương tự :\(\dfrac{1}{b+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)
\(\dfrac{1}{a+c+1}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân các bất đẳng thức trên lại với nhau về theo vế ,ta được:
\(\dfrac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{1}{8}\)
Dấu "=" xảy ra khi:\(a=b=c=\dfrac{1}{4}\)
Vậy giá trị lớn nhất của (a+b)(b+c)(c+a) là \(\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{4}\)