Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)
Tương tự:
\(\frac{1}{b^3(c+a)}+\frac{b(c+a)}{4}\geq \frac{1}{b}=ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq \frac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow \text{VT}+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}=3\)
\(\Rightarrow \text{VT}\ge \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Lời giải:
Đặt vế trái là $A$
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)(a+b+b+c+c+c)\geq (1+1+1+1+1+1)^2\)
\(\Leftrightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{36}{a+2b+3c}\)
Hoàn toàn TT:
\(\frac{1}{b}+\frac{2}{c}+\frac{3}{a}\geq \frac{36}{b+2c+3a}\)
\(\frac{1}{c}+\frac{2}{a}+\frac{3}{b}\geq \frac{36}{c+2a+3b}\)
Cộng theo vế:
\(\Rightarrow 6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 36A\)
\(\Rightarrow A\leq \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo đkđb: \(ab+bc+ac=abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Do đó: \(A\leq \frac{1}{6}< \frac{3}{16}\) (đpcm)
\(\frac{1}{a^2\left(b+c\right)}+\frac{1}{b^2\left(c+a\right)}+\frac{1}{c^2\left(a+b\right)}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\ge\frac{3}{2}\)( GT abc = 1 )
\(\Leftrightarrow\frac{bc}{ab+ac}+\frac{ac}{ab+ac}+\frac{ab}{ac+bc}\ge\frac{3}{2}\). Đặt \(\hept{\begin{cases}ab=x\\bc=y\\ac=z\end{cases}\left(x,y,z>0\right)}\)ta được bất đẳng thức Nesbitt quen thuộc :
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\)( em không chứng minh )
Vậy ta có đpcm
Đẳng thức xảy ra <=> x = y = z <=> a = b = c = 1
Do giả thiết abc=1abc=1 nên
\dfrac{1}{a^2\left(b+c\right)}=\dfrac{bc}{a^2bc\left(b+c\right)}=\dfrac{bc}{a\left(b+c\right)}=\dfrac{bc}{ab+ac}a2(b+c)1=a2bc(b+c)bc=a(b+c)bc=ab+acbc
Đặt x=bc,y=ca,z=abx=bc,y=ca,z=ab thì x,y,z>0x,y,z>0 và bất đẳng thức cần chứng minh trở thành bất đẳng thức quen thuộc
\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}y+zx+z+xy+x+yz≥23.
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( 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 \(\Leftrightarrow a=b=c=1\)