Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Khó quá. Đúng là Câu Hỏi Hay!!
a)Áp dụng BĐT AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
Nhân theo vế 2 BĐT trên có:
\(A\ge9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)
Khi \(a=b=c\)
Bài 2:
a)Sửa đề \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{\left(1+1\right)^2}{x+y}=\dfrac{4}{x+y}\)
Khi \(x=y\)
b)Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có:
\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{4}{2b}=\dfrac{2}{b}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c};\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\ge\dfrac{2}{a}\)
Cộng theo vế 3 BĐT trên ta có:
\(2VT\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2VP\Leftrightarrow VT\ge VP\)
Khi \(a=b=c\)
Câu 1: Với \(a;b;c>0\), theo bất đẳng thức Cauchy:
\(a+b+c\ge3.\sqrt[3]{abc}\). Dấu "=" xảy ra khi \(a=b=c\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3.\sqrt[3]{\dfrac{1}{abc}}\). Dấu "=" xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)
Nhân theo vế ta được \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
\(\Rightarrow MinA=9\)
Dấu "=" xảy ra khi a = b = c
\(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{1}{2}ab\)
Tương tự: \(\dfrac{b}{1+c^2}\ge b-\dfrac{1}{2}bc\) ; \(\dfrac{c}{1+a^2}\ge c-\dfrac{1}{2}ca\)
Cộng vế:
\(P\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{6}\left(a+b+c\right)^2=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)
A=\(\left(a+b\right)\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
= \(\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{a}{b}+\dfrac{b}{a}\)
= \(2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
Áp dụng BĐT cô si cho 2 số ta có
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)
⇔\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
⇔\(2+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge4\)
⇔ A ≥4
=> Min A =4
dấu "=" xảy ra khi
\(\dfrac{a}{b}=\dfrac{b}{a}\)
⇔a2=b2
⇔a=b
vậy Min A =4 khi a=b
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
a) Áp dụng BĐT bunhiacopxki ta có:
A= \(a^2+b^2\) \(\geq\) \(\dfrac{\left(a+b\right)^2}{2}=\dfrac{1}{2}\)
Vậy Min A= \(\dfrac{1}{2}\) khi a=b=\(\dfrac{1}{2}\)
b) Ta có: B= \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\)
\(\Leftrightarrow\) B= \(\left(\dfrac{a^2}{b}+b\right)+\left(\dfrac{b^2}{a}+a\right)-\left(a+b\right)\) \(\geq\) \(2\sqrt{\dfrac{a^2}{b}.b}+2\sqrt{\dfrac{b^2}{a}.a}-a-b\) = \(2a+2b-a-b\) \(=a+b=1\)
Từ đó suy ra: \(\dfrac{a^2}{b}+\dfrac{b^2}{a}\) \(\geq\) 1
Vậy Min B = 1 khi a=b=\(\dfrac{1}{2}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel ta có:
\(B=\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge\dfrac{\left(1+1+1\right)^2}{a+1+b+1+c+1}=\dfrac{9}{0+3}=3\)
Dấu "=" xảy ra khi a = b = c = 0
\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)
\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)
Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)
Lời giải:
Từ điều kiện
\(a+b+c=3abc\Rightarrow A=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=\frac{3(ab+bc+ac)}{a+b+c}(1)\)
Theo hệ quả của BĐT AM-GM:
\(a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)\)
\(\Leftrightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)=(a+b+c)^2\)
\(\Leftrightarrow ab+bc+ac\geq a+b+c(2)\)
Từ \((1),(2)\Rightarrow A\geq 3\)
Do đó \(A_{\min}=3\Leftrightarrow a=b=c=1\)