Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
\(a+b+c=abc\Rightarrow a(a+b+c)=a^2bc\)
\(\Rightarrow a(a+b+c)+bc=bc(a^2+1)\)
\(\Leftrightarrow (a+b)(a+c)=bc(a^2+1)\Rightarrow a^2+1=\frac{(a+b)(a+c)}{bc}\)
\(\Rightarrow \frac{1}{\sqrt{a^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}\)
Hoàn toàn tương tự với các phân thức còn lại
\(\Rightarrow \text{VT}=\frac{1}{\sqrt{a^2+1}}+\frac{1}{\sqrt{b^2+1}}+\frac{1}{\sqrt{c^2+1}}=\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ac}{(b+a)(b+c)}}+\sqrt{\frac{ab}{(c+a)(c+b)}}\)
Áp dụng BĐT Cauchy:
\(\sqrt{\frac{bc}{(a+b)(a+c)}}+\sqrt{\frac{ac}{(b+a)(b+c)}}+\sqrt{\frac{ab}{(c+a)(c+b)}}\leq \frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)+\frac{1}{2}\left(\frac{a}{b+a}+\frac{c}{b+c}\right)+\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)
\(=\frac{1}{2}\left(\frac{b+a}{b+a}+\frac{c+b}{c+b}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
\(\Rightarrow \text{VT}\leq \frac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=\sqrt{3}$
Từ \(abc+a+b=3ab\Leftrightarrow c+\dfrac{1}{a}+\dfrac{1}{b}=3\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b}\right)\rightarrow\left(x;y\right)\left(x;y>0\right)\Rightarrow c+x+y=3\)
BĐT cần chứng minh là:
\(\sqrt{\dfrac{1}{x+y+xy}}+\sqrt{\dfrac{1}{y+a+ay}}+\sqrt{\dfrac{1}{x+a+ax}}\ge\sqrt{3}\)
Áp dụng BĐT AM-GM ta có:
\(VT\ge3\sqrt[6]{\dfrac{1}{\left(x+y+xy\right)\left(x+a+ax\right)\left(a+y+ay\right)}}\ge\sqrt{3}\)
\(\Leftrightarrow (x+y+xy)(x+a+ax)(a+y+ay)\leq \frac{1}{27}\)
BĐT này luôn đúng vì ta có 2 BĐT phụ sau luôn đúng theo AM-GM \(mnp\le\left(\dfrac{m+n+p}{3}\right)^3;mn+np+mp\le\dfrac{\left(m+n+p\right)^2}{3}\)
Ok. Done ! 
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow VT\ge3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\)
Chứng minh \(3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\ge3\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\left(c+ab\right)\left(a+bc\right)\le\dfrac{\left(c+a+ab+bc\right)^2}{4}=\dfrac{\left[b\left(a+c\right)+c+a\right]^2}{4}=\dfrac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)
Thiết lập tương tự và thu lại ta có
\(\Rightarrow\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(a+1\right)^2\left(c+1\right)^2}{64}\)
\(\Rightarrow64\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(c+1\right)^2\left(a+1\right)^2\)
\(\Leftrightarrow8\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+1\right)\left(c+1\right)\left(a+1\right)\)
Cần chứng minh rằng \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)
Áp dụng bất đẳng thức Cauchy \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\dfrac{3+3}{3}\right)^3=8\left(đpcm\right)\)
\(\Rightarrowđpcm\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
Chứng minh rằng \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
\(\Leftrightarrow18\ge3\left(3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}\right)\)
\(\Leftrightarrow18\ge9+3\sqrt[3]{bc}+3\sqrt[3]{ca}+3\sqrt[3]{ab}\)
\(\Leftrightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a+b+1\ge3\sqrt[3]{ab}\\b+c+1\ge3\sqrt[3]{bc}\\c+a+1\ge3\sqrt[3]{ca}\end{matrix}\right.\)
\(\Rightarrow2\left(a+b+c\right)+3\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\)
\(\Rightarrow9\ge3\sqrt[3]{ab}+3\sqrt[3]{bc}+3\sqrt[3]{ca}\) ( đpcm )
Vì \(\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)
Mà \(\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{9}{3+\sqrt[3]{bc}+\sqrt[3]{ca}+\sqrt[3]{ab}}\)
\(\Rightarrow\dfrac{a^2}{a+\sqrt[3]{bc}}+\dfrac{b^2}{b+\sqrt[3]{ca}}+\dfrac{c^2}{c+\sqrt[3]{ab}}\ge\dfrac{3}{2}\)( đpcm )
Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
Từ giả thiết ta suy ra
\(\dfrac{1}{a}+\dfrac{1}{b}+c=3\)
Đặt \(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};c\right)\Rightarrow x+y+z=3\)
\(VT=\dfrac{1}{\sqrt{xy+x+y}}+\dfrac{1}{\sqrt{yz+y+z}}+\dfrac{1}{\sqrt{xz+x+z}}\)
Ta chứng minh: \(\left(x+1+y\right)^2\ge3\left(xy+x+y\right)\)(Luôn đúng)
\(\Rightarrow VT\ge\dfrac{\sqrt{3}}{x+y+1}+\dfrac{\sqrt{3}}{y+z+1}+\dfrac{\sqrt{3}}{z+x+1}\ge\dfrac{9\sqrt{3}}{2\left(x+y+z\right)+3}=\sqrt{3}\)
https://hoc24.vn/hoi-dap/tim-kiem?q=Cho+c%C3%A1c+s%E1%BB%91+th%E1%BB%B1c+d%C6%B0%C6%A1ng+a,+b,+c+tho%E1%BA%A3+m%C3%A3n:+abc+a+b=3ababc+a+b=3ababc+a+b=3ab.+Ch%E1%BB%A9ng+minh+r%E1%BA%B1ng:+%E2%88%9Aaba+b+1+%E2%88%9Abbc+c+1+%E2%88%9Aaca+c+1%E2%89%A5%E2%88%9A3aba+b+1+bbc+c+1+aca+c+1%E2%89%A53\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{b}{bc+c+1}}+\sqrt{\dfrac{a}{ca+c+1}}\ge\sqrt{3}&id=695796