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Bài 1:
Biểu thức chỉ có giá trị lớn nhất, không có giá trị nhỏ nhất.
\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{1}{x+1}+1-\frac{1}{y+1}+1-\frac{1}{z+1}\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Giờ chỉ cần cho biến $x$ nhỏ vô cùng đến $0$, khi đó giá trị biểu thức trong ngoặc sẽ tiến đến dương vô cùng, khi đó P sẽ tiến đến nhỏ vô cùng, do đó không có min
Nếu chuyển tìm max thì em tìm như sau:
Áp dụng BĐT Cauchy_Schwarz:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq \frac{(1+1+1)^2}{x+1+y+1+z+1}=\frac{9}{x+y+z+3}=\frac{9}{4}\)
Do đó: \(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\leq 3-\frac{9}{4}=\frac{3}{4}\)
Vậy \(P_{\min}=\frac{3}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
Bài 2:
Áp dụng BĐT Cauchy-Schwarz :
\(\frac{1}{a+3b+2c}=\frac{1}{9}\frac{9}{(a+c)+(b+c)+2b}\leq \frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
\(\Rightarrow \frac{ab}{a+3b+2c}\leq \frac{1}{9}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{a}{2}\right)\)
Hoàn toàn tương tự:
\(\frac{bc}{b+3c+2a}\leq \frac{1}{9}\left(\frac{bc}{b+a}+\frac{bc}{c+a}+\frac{b}{2}\right)\)
\(\frac{ac}{c+3a+2b}\leq \frac{1}{9}\left(\frac{ac}{c+b}+\frac{ac}{a+b}+\frac{c}{2}\right)\)
Cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{1}{9}\left(\frac{b(a+c)}{a+c}+\frac{a(b+c)}{b+c}+\frac{c(a+b)}{a+b}+\frac{a+b+c}{2}\right)\)
hay \(\text{VT}\leq \frac{a+b+c}{6}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
Áp dụng BĐT AM-Gm: ( dạng \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\))
\(VT=\sum\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\sum\dfrac{a}{2}+\sum\left[\dfrac{ab}{a+c}+\dfrac{bc}{a+c}\right]\right)\)
\(=\dfrac{1}{9}\left(\dfrac{a+b+c}{2}+a+b+c\right)=\dfrac{1}{6}\left(a+b+c\right)\)
\(\le\dfrac{1}{6}\sqrt{3\left(a^2+b^2+c^2\right)}=1\) (đpcm)
Dấu = xảy ra khi a=b=c=2
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}\leq 1(*)\)
\((*)\Rightarrow \frac{1}{a+1}=1-\frac{a}{a+1}\geq \frac{2b}{b+1}+\frac{3c}{c+1}=\frac{b}{b+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{c}{c+1}+\frac{c}{c+1}\geq 5\sqrt[5]{\frac{b^2c^3}{(b+1)^2(c+1)^3}}(1)\)
\((*)\Rightarrow \frac{1}{b+1}=1-\frac{b}{b+1}\geq \frac{a}{a+1}+\frac{b}{b+1}+\frac{3c}{c+1}=\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{c}{c+1}+\frac{c}{c+1}\geq 5\sqrt[5]{\frac{abc^3}{(a+1)(b+1)(c+1)^3}}(2)\)
\((*)\Rightarrow \frac{1}{c+1}=1-\frac{c}{c+1}\geq \frac{a}{a+1}+\frac{2b}{b+1}+\frac{2c}{c+1}=\frac{a}{a+1}+\frac{b}{b+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{c}{c+1}\geq 5\sqrt[5]{\frac{ab^2c^2}{(a+1)(b+1)^2(c+1)^2}}(3)\)
Lấy \((1).(2)^2.(3)^3\) rồi rút gọn ta suy ra \(ab^2c^3\leq \frac{1}{5^6}\)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{5}$
Ta có \(a\left(1-a\right)\left(1-b\right)\ge0\)
\(\Leftrightarrow a^2b\ge a^2+ab-a\)
Tương tự \(b^2c\ge b^2+bc-b;c^2a\ge c^2+ca-a\)
\(\Rightarrow a^2b+b^2c+c^2a+1\ge a^2+b^2+c^2+ab+bc+ca-a-b-c+1\)\(=a^2+b^2+c^2+\left(1-a\right)\left(1-b\right)\left(1-c\right)+abc\ge a^2+b^2+c^2\)
Hay \(a^2+b^2+c^2\le a^2b+b^2c+c^2a+1\)
Áp dụng BĐT Cosi cho 3 số dương ta có:
\(a+2b+3c\ge3\sqrt[3]{a.2b.3c}\)
\(\Rightarrow\left(a.2b.3c\right)^3\ge162abc\)
\(\Rightarrow abc\le\frac{1}{162}\)
\(a=2b=3c=\frac{1}{3}\)
Dấu " = " sảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{matrix}\right.\)
-2<=x+1<3
=>-3<=x<2
B=[-3;2)
\(x^2< =4\)
=>(x-2)(x+2)<=0
=>-2<=x<=2
=>A=[-2;2]
A\B={2}
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:
\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)
Đẳng thức xảy ra khi a=b=c=d
Bài 1:
Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)
Áp dụng bđt Cauchy Schwarz có:
\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)
Lại sử dụng bđt Cauchy schwarz ta có:
\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)
=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)
hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
Áp dụng bđt Cosi ta có:
\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)
Nhân các vế của 3 bđt trên ta đc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)
=> Đpcm
\(-1\le a,b,c\le4\) \(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)\left(a-4\right)\le0\\\left(b+1\right)\left(b-4\right)\le0\\\left(c+1\right)\left(c-4\right)\le0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2\le3a+4\\b^2\le3b+4\\c^2\le3c+4\end{matrix}\right.\Rightarrow\)\(\left\{{}\begin{matrix}a^2\le3a+4\\2b^2\le6b+8\\3c^2\le9c+12\end{matrix}\right.\)
Cộng vế theo vế \(\Rightarrow a^2+2b^2+3c^2\le3\left(a+2b+3c\right)+24\)
Thay \(a+2b+3c\le4\)
\(\Rightarrow a^2+2b^2+3c^2\le3.4+24=36\) (đpcm)
Happy women's day