Cho a,b > 0 và \(a^2+b^2\le2\) . Tìm max \(A=a\sqrt{3b\left(a+2b\right)}+b\sqrt{3a\left(b+2a\right)}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng BĐT Cauchy ta có : \(2\ge a^2+b^2\ge2\sqrt{a^2b^2}=2ab\Rightarrow ab\le1\)
Áp dụng BĐT Bunhiacopxki :
\(\left(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\right)^2\le\left(a^2+b^2\right)\left[3\left(a^2+b^2\right)+12ab\right]\)
\(\le2\left(3.2+12.1\right)=36\)
\(\Rightarrow a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\le6\)
Dấu "=" xảy ra khi a = b = 1
ÁP DỤNG BĐT CÔ SI ,TA CÓ:
\(\sqrt{3a\left(a+2b\right)}\le\frac{3a+\left(a+2b\right)}{2}=2a+b\)\(\Leftrightarrow a\sqrt{3a\left(a+2b\right)}\le a\left(2a+b\right)=2a^2+ab\left(1\right)\)
(VÌ a,b khong âm). C/M TƯƠNG TỰ TA CÓ \(b\sqrt{3b\left(b+2a\right)}\le2b^2+ab\left(2\right)\)
TA CÓ :\(2ab\le a^2+b^2\le2\left(3\right)\).TỪ (1),(2),(3) TA CÓ;
\(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\le2a^2+2b^2+ab+ab\le\)\(2\left(a^2+b^2\right)+2ab\le4+2=6\)
DẤU ĐẲNG THỨC XẢY RA KHI a=b=1
Áp dụng bất đẳng thức Cô-si :
\(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\le a\cdot\frac{3a+a+2b}{2}+b\cdot\frac{3b+b+2a}{2}\)
\(=a\cdot\frac{4a+2b}{2}+b\cdot\frac{4b+2a}{2}\)
\(=a\left(2a+b\right)+b\left(2b+a\right)\)
\(=2a^2+2b^2+2ab\)
\(=2\left(a^2+b^2+ab\right)\le2\left(2+\frac{a^2+b^2}{2}\right)=2\left(2+\frac{2}{2}\right)=6\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=1\)
p/s: có gì chiều giải nốt, giờ đi ăn cơm @@
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
\(\sqrt{3b\left(a+2b\right)}\le\frac{3b+\left(a+2b\right)}{2}\); \(\sqrt{3a\left(b+2a\right)}\le\frac{3a+\left(b+2a\right)}{2}\)
=> M\(\le a\frac{a+5b}{2}+b\frac{5a+b}{2}\)=\(\frac{a^2+b^2+10ab}{2}\)\(\le\frac{6\left(a^2+b^2\right)}{2}\)( áp dụng 2ab\(\le a^2+b^2\))=3(a2+b2)\(\le\)6
dấu = khi a =b =1
Ta có \(\sqrt{3b\left(a+2b\right)}\le\frac{1}{2}\left(3b+a+2b\right)=\frac{1}{2}\left(a+5b\right)\)
\(\sqrt{3a\left(b+2a\right)}\le\frac{1}{2}\left(5a+b\right)\)
=> \(P\le\frac{1}{2}\left(a^2+b^2+10ab\right)\)
Mà \(ab\le\frac{1}{2}\left(a^2+b^2\right)\le\frac{1}{2}.2=1\)
=> \(P\le\frac{1}{2}\left(2+10\right)=6\)
Vậy MaxP=6 khi a=b=1
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((a\sqrt{3a(a+2b)}+b\sqrt{3b(b+2a)})^2\leq (a^2+b^2)[3a(a+2b)+3b(b+2a)]\)
\((a\sqrt{3a(a+2b)}+b\sqrt{3b(b+2a)})^2\leq (a^2+b^2)(3a^2+3b^2+12ab)\)
Theo BĐT Cô-si: \(a^2+b^2\geq 2ab\Rightarrow 12ab\leq 6(a^2+b^2)\)
Do đó:
\((a\sqrt{3a(a+2b)}+b\sqrt{3b(b+2a)})^2\leq (a^2+b^2)(3a^2+3b^2+6a^2+6b^2)=9(a^2+b^2)^2\)
Mà \(a^2+b^2\leq 2\)
\(\Rightarrow (a\sqrt{3a(a+2b)}+b\sqrt{3b(b+2a)})^2\leq 9.2^2=36\)
\(\Rightarrow a\sqrt{3a(a+2b)}+b\sqrt{3b(b+2a)}\leq \sqrt{36}=6\)
(đpcm)
Dấu bằng xảy ra khi $a=b=1$
Theo bất đẳng thức Bunhiacopxki, ta có :
\(P\le\sqrt{\left[\left(\sqrt{a}\right)^2+\left(\sqrt{b}\right)^2\right]\left[\left(\sqrt{b+1}\right)^2+\left(\sqrt{a+1}\right)^2\right]}\)
\(=\sqrt{\left(a+b\right)\left(a+b+2\right)}\)
\(\Rightarrow P\le\sqrt{2\left(2+2\right)}=2\sqrt{2}\)
Vậy : GTLN của P là \(2\sqrt{2}\). Dấu đẳng thức xảy ra khi và chỉ khi \(a=b=1\)
b.\(ĐK:x;y\in Z^+;x;y\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)
\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)
\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)
\(\Leftrightarrow x=\dfrac{5y}{y-5}\)
\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )
Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)
TH1:
\(y-5=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm ) ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )
Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:
\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)
Câu a mik ko bt nên bạn tham khảo nhé:
https://hoc24.vn/cau-hoi/cho-a-b-c-0-va-day-ti-so-dfrac2bc-aadfrac2c-babdfrac2ab-cctinh-p-dfracleft3a-2brightleft3b-2crightleft.177725456910
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)
Áp dụng BĐT AM - GM, ta có:
\(2\ge a^2+b^2\ge2ab\)
\(\Leftrightarrow ab\le1\)
\(A=a\sqrt{3b\left(a+2b\right)}+b\sqrt{3a\left(b+2a\right)}\)
\(\le\dfrac{a\left(3b+a+2b\right)}{2}+\dfrac{b\left(3a+b+2a\right)}{2}\)
\(=\dfrac{a\left(5b+a\right)+b\left(5a+b\right)}{2}\)
\(=\dfrac{a^2+10ab+b^2}{2}\)
\(\le\dfrac{2+10}{2}=6\)
Dấu "=" xảy ra khi a = b = 1