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Câu 2 : a,b,c > 0. CM : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Giải :
C1 : Áp dụng bất đẳng thức Cauchy - Schwarz dạng Engel ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\left(ĐPCM\right)\)
Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\).
C2 : Đầy đủ hơn với cách giải đúng của bạn Hoàng Thiên Di :
Áp dụng BĐT AM-GM cho 3 số dương (sgk là cosi :v)
\(a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\ge3+2+2+2=9\left(ĐPCM\right)\)
Câu 3 : a,b,c > 0. CM : \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
Giải :
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
\(\Leftrightarrow\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\)
Theo bất đẳng thức Cosi : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{yx}}=2\)
Thay vào các vế được : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\sqrt{1}=2\)
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\sqrt{1}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\sqrt{1}=2\)
\(\Leftrightarrow2+2+2\ge6\) (đúng)
BĐT được c/m.
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge6\)
=> \(-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6\)
=> \(-\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le-6.\frac{3}{2}\)
=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
=> \(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\)
=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(1)
Dễ thấy \(\frac{a}{b}+\frac{b}{a}\ge2\)(với a,b > 0)
=> (1) đúng
=> BĐTđược chứng minh
b)Đặt \(A=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(a,b,c>0\right)\).
\(A=4\left(a+b+c\right)-3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\).
\(A=\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\).
Vì \(a>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(4a+\frac{1}{a}\ge2\sqrt{4.a.\frac{1}{a}}=4\left(1\right)\).
Dấu bằng xảy ra \(\Leftrightarrow4a=\frac{1}{a}\Leftrightarrow a=\frac{1}{2}\).
Chứng minh tương tự, ta được:
\(4b+\frac{1}{b}\ge4\left(b>0\right)\left(2\right)\).
Dấu bằng xảy ra \(\Leftrightarrow b=\frac{1}{2}\).
Chứng minh tương tự, ta được:
\(4c+\frac{1}{c}\ge4\left(c>0\right)\left(3\right)\).
Dấu bằng xảy ra \(\Leftrightarrow c=\frac{1}{2}\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)\ge4+4+4=12\).
\(\Leftrightarrow\left(4a+\frac{1}{a}\right)+\left(4b+\frac{1}{b}\right)+\left(4c+\frac{1}{c}\right)-3\left(a+b+c\right)\ge\)\(12-3\left(a+b+c\right)\).
\(\Leftrightarrow A\ge12-3\left(a+b+c\right)\left(4\right)\).
Mặt khác, ta có: \(a+b+c\le\frac{3}{2}\).
\(\Leftrightarrow3\left(a+b+c\right)\le\frac{9}{2}\).
\(\Rightarrow-3\left(a+b+c\right)\ge-\frac{9}{2}\).
\(\Leftrightarrow12-3\left(a+b+c\right)\ge\frac{15}{2}\left(5\right)\).
Dấu bằng xảy ra \(\Leftrightarrow a+b+c=\frac{3}{2}\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(A\ge\frac{15}{2}\).
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\).
Vậy với \(a,b,c>0\)và \(a+b+c\le\frac{3}{2}\)thì \(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{15}{2}\).
Bài 1:Cách thông thường nhất là sos hoặc cauchy-Schwarz nhưng thôi ko làm:v Thử cách này cho nó mới dù rằng ko chắc
Giả sử \(a\ge b\ge c\Rightarrow c\le1\Rightarrow a+b=3-c\ge2\) và \(a\ge1\)
Ta có \(LHS=a^3.a+b^3.b+c^3.c\)
\(=\left(a^3-b^3\right)a+\left(b^3-c^3\right)\left(a+b\right)+c^3\left(a+b+c\right)\)
\(\ge\left(a^3-b^3\right).1+\left(b^3-c^3\right).2+3c^3\)
\(=a^3+b^3+c^3=RHS\)
Đẳng thức xảy ra khi a = b = c = 1
Áp dụng BĐT Cauchy – Schwarz, ta được:
\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\ge\frac{\left(a+b+c\right)^3}{b+c+a+c+a+b}\)
\(=\frac{\left(a+b+c\right)^3}{2\left(a+b+c\right)}=\frac{\left(a+b+c\right)^2}{2}\ge\frac{a^2+b^2+c^2}{2}=\frac{1}{2}\)
ミ★长 - ƔξŦ★彡vãi cả cauchy-schwarz cho bậc 3: \("\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{\left(a+b+c\right)^3}{b+c+c+a+a+b}\)
Thiết nghĩ nên sửa đề \(a,b,c>0\) thôi chứ là gì có d? Mà nếu a >b >c > d > 0 thì liệu dấu = có xảy ra?
Áp dụng BĐT Cauchy-Scwarz ta có: \(LHS\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{a^2+b^2+c^2}{2}=\frac{1}{2}\)
BĐT Bunhiacopxky em chưa học cô ạ
Cô cong cách nào không ạ
Nguyễn Thị Nguyệt Ánh:
Vậy thì bạn có thể chứng minh $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$ thông qua BĐT Cô-si:
Áp dụng BĐT Cô-si:
$x+y+z\geq 3\sqrt[3]{xyz}$
$xy+yz+xz\geq 3\sqrt[3]{x^2y^2z^2}$
Nhân theo vế:
$(x+y+z)(xy+yz+xz)\geq 9xyz$
$\Rightarrow \frac{xy+yz+xz}{xyz}\geq \frac{9}{x+y+z}$
hay $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Từ gt =>
\(\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)= \(\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\)\(\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
( Theo Cô-si )
Vậy :
\(\left\{{}\begin{matrix}\frac{1}{1+a}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\ge0\\\frac{1}{1+b}\ge3\sqrt[3]{\frac{cda}{\left(1+c\right)\left(1+d\right)\left(1+a\right)}}\ge0\\\frac{1}{1+c}\ge3\sqrt[3]{\frac{dca}{\left(1+d\right)\left(1+c\right)\left(1+a\right)}}\ge0\\\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\ge0\end{matrix}\right.\)
=> \(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge81\frac{abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\Rightarrow abcd\le\frac{1}{81}\)