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1) Áp dụng bất đẳng Bunyakovsky dạng cộng mẫu ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)
\(=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)
(Cauchy 3 số) Dấu "=" xảy ra khi: a = b = c
2) Áp dụng kết quả phần 1 ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^2+c^3\right)^2}{3\cdot\frac{1}{3}}=\left(a^3+b^3+c^3\right)^2\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
Ta có: \(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\) và \(\left\{{}\begin{matrix}a^2+1\ge2a\\b^2+1\ge2b\\c^2+1\ge2c\end{matrix}\right.\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)+3\ge2\left(a+b+c+ab+bc+ca\right)=12\)
\(\Rightarrow a^2+b^2+c^2\ge3\left(1\right)\)
Ta lại có:
\(\left\{{}\begin{matrix}\dfrac{a^3}{b}+ab\ge2a^2\\\dfrac{b^3}{c}+bc\ge2b^2\\\dfrac{c^3}{a}+ca\ge2c^2\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-ab-bc-ca\ge a^2+b^2+c^2\left(2\right)\)
Từ (1) và (2) \(\RightarrowĐPCM\)
Cách 1
Áp dụng bđt Cauchy ta có
\(\frac{a^3}{b}+b+1\ge3a,\frac{b^3}{c}+c+1\ge3b,\frac{c^3}{a}+a+1\ge3a\)
Cộng từng vế 3 bđt trên ta có
\(A=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a+b+c\right)-3\)
Mặt khác (a+b+c)2+3(a+b+c)\(\ge\)18 (biến đổi tương đương là c/m được)
Đặt m=a+b+c
=> t2+3t-18\(\ge\)0
=> t\(\ge\)3
=> A\(\ge\)3
Dấu "=" xảy ra khi a=b=c=1
Cách 2,rất phức tạp :(
\(6=a+b+c+ab+bc+ca\le\frac{\left(a+b+c\right)^2+3\left(a+b+c\right)}{3}\)
Suy ra \(\left(a+b+c\right)^2+3\left(a+b+c\right)-18\ge0\)
\(\Leftrightarrow a+b+c\ge3\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge9\).
Mà \(VT\le3\left(a^2+b^2+c^2\right)\Rightarrow3\left(a^2+b^2+c^2\right)\ge9\Leftrightarrow a^2+b^2+c^2\ge3\)
Ta chứng minh BĐT sau = sos cho đẹp: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
\(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{b}-\frac{a^2b}{b}\right)\ge0\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)}{b}-\Sigma_{cyc}a\left(a-b\right)+\Sigma_{cyc}a\left(a-b\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2\left(a-b\right)^2}{b}+\frac{1}{2}\left(a-b\right)^2\ge0\Leftrightarrow\left(a-b\right)^2\left(\frac{a^2}{b}+\frac{1}{2}\right)\ge0\) (đúng)
Do vậy: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\ge3^{\left(đpcm\right)}\)
Xảy ra đẳng thức 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 \(A_{min}=3\Leftrightarrow a=b=c=2\)
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:
\(\sum\frac{a^2}{a+\sqrt[3]{bc}}\geq\sum\frac{a^2}{a+\frac{b+c+1}{3}}=\sum\frac{9a^2}{3(3a+b+c)+a+b+c}\)
\(=\sum\frac{9a^2}{10a+4b+4c}\geq\frac{9(a+b+c)^2}{(10a+4b+4c)}=\frac{9(a+b+c)^2}{18(a+b+c)}=\frac{3}{2}\)
Ta có: \(\dfrac{ab}{c+1}=\dfrac{ab}{b+c+a+c}\le\dfrac{1}{4}\left(\dfrac{ab}{b+c}+\dfrac{ab}{a+c}\right)\)
Tương tự cho 2 BĐT còn lại:
\(\dfrac{bc}{a+1}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{b+1}\le\dfrac{1}{4}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế các BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
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
Đặt vế trái là P
\(P=\dfrac{1.c+ab}{a+b}+\dfrac{1.a+bc}{b+c}+\dfrac{1.b+ac}{a+c}=\dfrac{c\left(a+b+c\right)+ab}{a+b}+\dfrac{a\left(a+b+c\right)+bc}{b+c}+\dfrac{b\left(a+b+c\right)+ac}{a+c}\)
\(P=\dfrac{ac+c^2+bc+ab}{a+b}+\dfrac{a^2+ac+ab+bc}{b+c}+\dfrac{ab+ac+b^2+bc}{a+c}\)
\(P=\dfrac{c\left(a+c\right)+b\left(a+c\right)}{a+b}+\dfrac{a\left(a+c\right)+b\left(a+c\right)}{b+c}+\dfrac{a\left(b+c\right)+b\left(b+c\right)}{a+c}\)
\(P=\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\)
Áp dụng BĐT Cô-si:
\(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)}}=2\left(a+c\right)\) (1)
Tương tự: \(\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(b+c\right)\) (2)
\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}\ge2\left(a+b\right)\) (3)
Cộng vế với vế (1);(2);(3):
\(2.\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+2.\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+2.\dfrac{\left(a+c\right)\left(b+c\right)}{a+b}\ge2\left(a+b\right)+2\left(b+c\right)+2\left(c+a\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(a+b\right)\left(b+c\right)}{a+c}+\dfrac{\left(a+c\right)\left(b+c\right)}{a+c}\ge2\left(a+b+c\right)=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)