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Đặt \(x=2a\)và \(y=2b\)suy ra \(\hept{\begin{cases}x>0\\y>0\\x+y\le2\end{cases}}\)
Suy ra : \(A=\frac{x}{y+2}+\frac{y}{x+2}+\frac{2}{x+y}\)
\(\Rightarrow A=\frac{x^2}{xy+2x}+\frac{y^2}{xy+2y}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(xy+x+y\right)}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(\frac{\left(x+y\right)^2}{4}+\left(x+y\right)\right)}+\frac{2}{x+y}\)
Đặt \(t=x+y\)( \(0< t\le2\))
Suy ra :
\(\Rightarrow A\ge\frac{t^2}{\frac{t^2}{2}+2t}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{4}{3}.\frac{1}{t}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{2t}{t+4}.\frac{4}{3}.\frac{1}{t}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3\left(t+4\right)}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3.\left(2+4\right)}}+\frac{2}{3}.\frac{1}{2}=\frac{5}{3}\)
"=" xảy ra khi \(x=y=\frac{1}{2}\)
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 \(A_{min}=3\Leftrightarrow a=b=c=2\)
\(\frac{1}{a^2+b^2+2}+\frac{1}{c^2+b^2+2}+\frac{1}{a^2+c^2+2}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{a^2+b^2}{a^2+b^2+2}+\frac{b^2+c^2}{b^2+c^2+2}+\frac{c^2+a^2}{c^2+a^2+2}\ge\frac{3}{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT\ge\frac{\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)^2}{2\left(a^2+b^2+c^2\right)+6}\)
\(\ge\frac{\sqrt{3\left(a^2b^2+b^2c^2+c^2a^2\right)}+2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)
\(\ge\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\)
Cần chứng minh \(\frac{2\left(a^2+b^2+c^2\right)+ab+bc+ca}{a^2+b^2+c^2}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge0\) *luôn đúng*
\(R=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{1}=9\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
Vậy GTNN của \(R\) là \(9\) khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt ~
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:
\(VT\le\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\)
Đặt \(A=\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)
Vì \(a,b,c>0\)nên áp dụng bất đẳng thức Cô-si cho 3 số dương, ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt[3]{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right).64}}\)\(=3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\left(1\right)\)
Chứng minh tương tự, ta được:
\(\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{1+c}{8}+\frac{1+a}{8}\ge\frac{3b}{4}\left(2\right)\)
\(\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3a}{4}\left(3\right)\)
Từ (1), (2), (3), ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\)\(+\frac{1+a}{8}+\frac{1+b}{8}+\frac{1+c}{8}+\frac{1+a}{8}+\frac{1+b}{8}+\frac{1+c}{8}\)\(\ge\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\)
\(\Leftrightarrow A+\frac{1+a}{4}+\frac{1+b}{4}+\frac{1+c}{4}\ge\frac{3a}{4}+\frac{3b}{4}+\frac{3c}{4}\)
\(\Leftrightarrow A+\frac{1+a+1+b+1+c}{4}\ge\frac{3a+3b+3c}{4}\)
\(\Leftrightarrow A+\frac{3+a+b+c}{4}\ge\frac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow A\ge\frac{3\left(a+b+c\right)}{4}-\frac{3-a-b-c}{4}\)
\(\Leftrightarrow A\ge\frac{3\left(a+b+c\right)-\left(a+b+c\right)}{4}-\frac{3}{4}\)
\(\Leftrightarrow A\ge\frac{2\left(a+b+c\right)}{4}-\frac{3}{4}\left(4\right)\)
Mặt khác, vì \(a,b,c>0\)nên áp dụng bất đẳng thức Cô-si cho 3 số dương, ta được:
\(a+b+c\ge3\sqrt[3]{abc}\)
Mà \(abc\ge1\Leftrightarrow\sqrt[3]{abc}\ge1\Leftrightarrow3\sqrt[3]{abc}\ge3\)
Do đó:
\(a+b+c\ge3\)
\(\Leftrightarrow2\left(a+b+c\right)\ge6\)
\(\Leftrightarrow\frac{2\left(a+b+c\right)}{4}\ge\frac{6}{4}=\frac{3}{2}\)
\(\Leftrightarrow\frac{2\left(a+b+c\right)}{4}-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\left(5\right)\)
Từ (4) và (5), ta được:
\(A\ge\frac{3}{4}\)(điều phải chứng minh)
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}a=b=c>0\\abc=1\end{cases}}\Leftrightarrow a=b=c=1\)
Vậy \(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\)với \(a,b,c>0\)và \(abc\ge1\)
Với x,y>0x,y>0 đã cho, áp dụng bất đẳng thức Cô si ta có
\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{x}+\dfrac{1+c}{y}\ge\dfrac{3a}{\sqrt[3]{xy}}(1+b)(1+c)a3+x1+b+y1+c≥3xy3a
Kỳ vọng rằng bất đẳng thức cần chứng minh trở thành đẳng thức khi a=b=c=1a=b=c=1, ta chọn x>0x>0 sao cho \dfrac{a^3}{\left(1+b\right)\left(1+c\right)}=\dfrac{1+b}{x}=\dfrac{1+c}{y}(1+b)(1+c)a3=x1+b=y1+c xảy ra khi a=b=c=1a=b=c=1, tức là \dfrac{1}{4}=\dfrac{2}{x}=\dfrac{2}{y}\Leftrightarrow x=y=841=x2=y2⇔x=y=8. Vì vậy
\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge\dfrac{3a}{4}(1+b)(1+c)a3+81+b+81+c≥43a
Viết hai bất đẳng thức tương tự rồi cộng theo vế ba bất đẳng thức này ta có
\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{3}{4}+\dfrac{a+b+c}{4}\ge(1+b)(1+c)a3+(1+c)(1+a)b3+(1+a)(1+b)c3+43+4a+b+c≥
\dfrac{3}{4}\left(a+b+c\right)43(a+b+c)
Hay \dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\dfrac{1}{2}\left(a+b+c\right)-\dfrac{3}{4}(1+b)(1+c)a3+(1+c)(1+a)b3+(1+a)(1+b)c3≥21(a+b+c)−43
Mà a+b+c\ge3\sqrt[3]{abc}\ge3a+b+c≥33abc≥3 . Suy ra
\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\dfrac{3}{4}(1+b)(1+c)a3+(1+c)(1+a)b3+(1+a)(1+b)c3≥43
Áp dụng BĐT cosi:
\(\left(a+\dfrac{1}{a}\right)\left(b+\dfrac{1}{b}\right)4=ab+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{1}{ab}\\ \ge ab+\dfrac{1}{ab}+2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=ab+\dfrac{1}{ab}+2\)
Áp dụng tiếp BĐT cosi:
\(ab+\dfrac{1}{ab}=\left(16ab+\dfrac{1}{ab}\right)-15ab\\ \ge2\sqrt{\dfrac{16ab}{ab}}-15ab=8-15ab\\ \ge8-15\cdot\dfrac{a+b}{4}=8-15\cdot\dfrac{1}{4}=\dfrac{17}{4}\)
\(\Leftrightarrow ab+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{1}{ab}\ge\dfrac{17}{4}+2=\dfrac{25}{4}\)
Dấu \("="\Leftrightarrow a=b=\dfrac{1}{2}\)