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gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Sửa lại đề là tìm Max nhé m.n
Ta có:
\(\frac{ab+bc+ca+6\left(a+b+c\right)+27}{\left(a+3\right)\left(b+3\right)\left(c+3\right)}=\frac{3}{5}\)
\(\Leftrightarrow\frac{\left(b+3\right)\left(c+3\right)+\left(c+3\right)\left(a+3\right)+\left(a+3\right)\left(b+3\right)}{\left(a+3\right)\left(b+3\right)\left(c+3\right)}=\frac{3}{5}\)
\(\Leftrightarrow\frac{5}{a+3}+\frac{5}{b+3}+\frac{5}{c+3}=3\Leftrightarrow\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}=0\)
Xét biểu thức:
\(\frac{a^2-4}{a^2-9}=\frac{\left(a-2\right)\left(a+2\right)}{\left(a-3\right)\left(a+3\right)}=\frac{a-2}{a+3}.\frac{a+2}{a-3}\)
tưởng tự:
\(\frac{b^2-4}{b^2-9}=\frac{b-2}{b+3}.\frac{b+2}{b-3},\frac{c^2-4}{c^2-9}=\frac{c-2}{c+3}.\frac{c+2}{c-3}\)
\(\Rightarrow\frac{a^2-4}{a^2-9}+\frac{b^2-4}{b^2-9}+\frac{c^2-4}{c^2-9}=\frac{a-2}{a+3}.\frac{a+2}{a-3}+\frac{b-2}{b+3}.\frac{b+2}{b-3}+\frac{c-2}{c+3}.\frac{c+2}{c-3}\)
Do vai trò của a và b và c như nhau nên ta giả sử
\(a\ge b\ge c\)
Khi đó ta có:
\(\frac{a-2}{a+3}\ge\frac{b-2}{b+3}\ge\frac{c-2}{c+3},\frac{a+2}{a-3}\le\frac{b+2}{b-3}\le\frac{c+2}{c-3}\)
Áp dụng bất đẳng thức chebyshev cho 2 bộ ngược chiều trên ta có
\(\frac{a-2}{a+3}.\frac{a+3}{a-2}+\frac{b-2}{b+3}.\frac{b+2}{b-3}+\frac{c-2}{c+3}.\frac{c+2}{c-3}\le\left(\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}\right).\left(\frac{a+2}{a-3}+\frac{b+2}{b-3}+\frac{c+2}{c-3}\right)\)
Mà \(\frac{a-2}{a+3}+\frac{b-2}{b+3}+\frac{c-2}{c+3}=0\)
\(\Rightarrow\frac{a^2-4}{a^2-9}+\frac{b^2-4}{b^2-9}+\frac{c^2-4}{c^2-9}\le0\)
\(\Rightarrow\frac{5}{a^2-9}+\frac{5}{b^2-9}+\frac{5}{c^2-9}\le-3\Rightarrow\frac{1}{a^2-9}+\frac{1}{b^2-9}+\frac{1}{c^2-9}\le\frac{-3}{5}\)
Dấu bằng xảy ra khi a=b=c=2
Để ý: \(ab+bc+ca=\frac{\left[\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)\right]}{2}\).
Do đó đặt \(a^2+b^2+c^2=x>0;a+b+c=y>0\). Bài toán được viết lại thành:
Cho \(y^2+5x=24\), tìm max:
\(P=\frac{x}{y}+\frac{y^2-x}{2}=\frac{5x}{5y}+\frac{y^2-x}{2}\)
\(=\frac{24-y^2}{5y}+\frac{y^2-\frac{24-y^2}{5}}{2}\)
\(=\frac{24-y^2}{5y}+\frac{3\left(y^2-4\right)}{5}\)\(=\frac{3y^3-y^2-12y+24}{5y}\)
Đặt \(y=t\). Dễ thấy \(12=3\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)=3t^2-5\left(ab+bc+ca\right)\)
Và dễ dàng chứng minh \(ab+bc+ca\le3\)
Suy ra \(3t^2=12+5\left(ab+bc+ca\right)\le27\Rightarrow t\le3\). Mặt khác do a, b, c>0 do đó \(0< t\le3\).
Ta cần tìm Max P với \(P=\frac{3t^3-t^2-12t+24}{5t}\)và \(0< t\le3\)
Ta thấy khi t tăng thì P tăng. Do đó P đạt giá trị lớn nhất khi t lớn nhất.
Khi đó P = 3. Vậy...
Ta có \(\sqrt{a^2-ab+b^2}=\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{1}{4}\left(a+b\right)^2}=\frac{1}{2}\left(a+b\right)\)
=> \(\frac{1}{\sqrt{a^2-ab+b^2}}\le\frac{1}{\frac{1}{2}\left(a+b\right)}=\frac{2}{a+b}\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
Chứng minh tương tự, rồi cộng lại, ta có
A\(\le\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
dấu = xảy ra <=> a=b=c=1
^_^
Có: \(9=\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow3\ge ab+bc+ca\)
Từ đây: \(D=\Sigma_{cyc}\frac{ab}{\sqrt{c^2+3}}\le\Sigma_{cyc}\frac{ab}{\sqrt{c^2+ab+bc+ca}}\)
\(=\Sigma_{cyc}\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}=\Sigma_{cyc}\sqrt{\frac{ab}{a+c}}.\sqrt{\frac{ab}{b+c}}\le\Sigma_{cyc}\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)
\(=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
Áp dụng bđt : x^2+y^2+z^2 >= (x+y+z)^2/3 ta có :
\(\frac{\sqrt{b^2+2a^2}}{ab}\)= \(\frac{\sqrt{a^2+b^2+a^2}}{ab}\)>= \(\frac{\sqrt{\frac{\left(a+b+a\right)^2}{3}}}{ab}\) = \(\frac{2a+b}{\sqrt{3}ab}\) = \(\frac{2}{\sqrt{3}b}+\frac{1}{\sqrt{3}a}\)
Tương tự : \(\frac{\sqrt{c^2+2b^2}}{bc}\)>= \(\frac{2}{\sqrt{3}c}+\frac{1}{\sqrt{3}b}\) ; \(\frac{\sqrt{a^2+2c^2}}{ac}\)>= \(\frac{2}{\sqrt{3}a}+\frac{1}{\sqrt{3}c}\)
=> \(\frac{\sqrt{b^2+2a^2}}{ab}\)+ \(\frac{\sqrt{c^2+2b^2}}{bc}\)+ \(\frac{\sqrt{a^2+2c^2}}{ac}\)>= \(\frac{3}{\sqrt{3}a}+\frac{3}{\sqrt{3}b}+\frac{3}{\sqrt{3}c}\)
= \(\frac{3}{\sqrt{3}}\).(1/a+1/b+1/c) = \(\sqrt{3}\).(ab+bc+ca)/abc = \(\sqrt{3}\).abc/abc = \(\sqrt{3}\)
Dấu "=" xảy ra <=> a=b=c=3
=> ĐPCM
k mk nha
1.Ta có: \(c+ab=\left(a+b+c\right)c+ab\)
\(=ac+bc+c^2+ab\)
\(=a\left(b+c\right)+c\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\)
CMTT \(a+bc=\left(c+a\right)\left(b+c\right)\)
\(b+ca=\left(b+c\right)\left(a+b\right)\)
Từ đó \(P=\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(c+a\right)\left(a+b\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(a+b\right)}}\)
Ta có: \(\sqrt{\frac{ab}{\left(a+b\right)\left(b+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}\right)\)( theo BĐT AM-GM)
CMTT\(\Rightarrow P\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{a+b}\right)\)
\(\Rightarrow P\le\frac{1}{2}.3\)
\(\Rightarrow P\le\frac{3}{2}\)
Dấu"="xảy ra \(\Leftrightarrow a=b=c\)
Vậy /...
\(\frac{a+1}{b^2+1}=a+1-\frac{ab^2-b^2}{b^2+1}=a+1-\frac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\frac{b^2\left(a+1\right)}{2b}\)
\(=a+1-\frac{b\left(a+1\right)}{2}=a+1-\frac{ab+b}{2}\)
Tương tự rồi cộng lại:
\(RHS\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}\)
\(\ge a+b+c+3-\frac{\frac{\left(a+b+c\right)^2}{3}+a+b+c}{2}=3\)
Dấu "=" xảy ra tại \(a=b=c=1\)
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\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+bc+ca+ab}}+\frac{c}{\sqrt{c^2+ca+ab+bc}}\)
\(=\frac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\sqrt{b}\cdot\sqrt{b}}{\sqrt{\left(b+c\right)\left(a+b\right)}}+\frac{\sqrt{c}\cdot\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(=\frac{\sqrt{a}}{\sqrt{a+b}}\cdot\frac{\sqrt{a}}{\sqrt{c+a}}+\frac{\sqrt{b}}{\sqrt{b+c}}\cdot\frac{\sqrt{b}}{\sqrt{a+b}}+\frac{\sqrt{c}}{\sqrt{c+a}}\cdot\frac{\sqrt{c}}{\sqrt{c+b}}\)
\(\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}+\frac{\frac{b}{b+c}+\frac{b}{a+b}}{2}+\frac{\frac{c}{c+a}+\frac{c}{b+c}}{2}\)
\(=\frac{\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}}{2}=\frac{3}{2}\)
Vậy Max A = 3/2 khi a = b = c = 1. (Max not Min)
làm như giỏi lắm í, thôi khỏi nói cũng biết, ko cần thể hiện đâu
\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a+b+c}{\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}}\)
Ta có: \(\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}\)
\(=\sqrt{ab+bc+ac+a^2}+\sqrt{ab+bc+ac+b^2}+\sqrt{ab+bc+ca+c^2}\)
\(=\sqrt{b\left(a+c\right)+a\left(a+c\right)}+\sqrt{b\left(a+b\right)+c\left(a+b\right)}+\sqrt{b\left(a+c\right)+c\left(a+c\right)}\)
\(=\sqrt{\left(a+c\right)\left(a+b\right)}+\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(\le\frac{a+c+a+b}{2}+\frac{a+b+b+c}{2}+\frac{a+c+b+c}{2}\)
\(\le\frac{2a+a+2b+b+2c+c}{2}=\frac{3a+3b+3c}{2}=\frac{3}{2}\left(a+b+c\right)\)
Suy ra : \(A=\frac{a+b+c}{\sqrt{3+a^2}+\sqrt{3+b^2}+\sqrt{3+c^2}}\ge\frac{2}{3}\)
Dấu "=" xảy ra khi và chỉ khi a=b=c=0
Vậy Amin = \(\frac{2}{3}\)
Chắc sai. Mong bạn giúp đỡ. Cảm ơn!
Hình như đề là tìm min mới đúng chứ Incursion_03 ? nếu tìm max khúc cuối bđt nó sẽ đổi chiều thế này:
* Nếu là tìm max
Ta có: \(\left(a^2+b^2+c^2\right)\left(b^2+c^2+a^2\right)\ge\left(ab+bc+ca\right)^2=3^2=9\) (BĐT Bunhiaxcopki)
Hay \(\left(a^2+b^2+c^2\right)^2\ge9\Leftrightarrow a^2+b^2+c^2\ge3\)
Mặt khác,ta lại có:
\(A^2=\frac{a^2}{3+a^2}+\frac{b^2}{3+b^2}+\frac{c^2}{3+c^2}\)
\(=\left(1+1+1\right)-\left(\frac{3}{3+a^2}+\frac{3}{3+b^2}+\frac{3}{3+c^2}\right)\)
\(=3-3\left(\frac{1}{3+a^2}+\frac{1}{3+b^2}+\frac{1}{3+c^2}\right)\)
\(\le3-\frac{27}{9+a^2+b^2+c^2}\ge3-\frac{27}{9+3}=\frac{3}{4}?!?\)
Suy ra \(A_{max}=\frac{\sqrt{3}}{2}?!?\)
Hồi nữa tui đăng bài tìm min lên sau.
Hay là bài nãy t làm sai?
Từ \(ab+bc+ca=3\Rightarrow a^2+b^2+c^2\ge3\) (biến đổi như bài trước)
Đặt \(a^2+b^2+c^2=3+m\left(m\ge0\right)\)
\(A^2=\frac{a^2}{3+a^2}+\frac{b^2}{3+a^2}+\frac{c^2}{3+c^2}\)
\(=\left(3+3+3\right)-\left(\frac{9}{3+a^2}+\frac{9}{3+b^2}+\frac{9}{3+c^2}\right)\)
\(\le9-\frac{\left(3+3+3\right)^2}{9+a^2+b^2+c^2}=9-\frac{81}{9+a^2+b^2+c^2}\)
\(=9-\frac{81}{12+m}=\frac{27+9m}{12+m}\ge\frac{9}{4}?!?\)"vẫn thế,không thay đổi được gì"
Êy !! Làm này giờ mới nhận ra mình bình phương sai -> bài làm sai. Bỏ mấy bài nãy đi bạn nhé!
Có lẽ là BĐT Cô-si :)
Hazzz, tìm max nhé ! Bài này là đề của sở nên ko sai đề đc :) t ngồi trong phòng thi nghĩ bài này trong 5' only !
Nhìn nhé !
\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a}{\sqrt{ab+bc+ca+a^2}}+\frac{b}{\sqrt{ab+bc+ca+b^2}}+\frac{c}{\sqrt{ab+bc+ca+c^2}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\frac{a}{a+b}}.\sqrt{\frac{a}{a+c}}+\sqrt{\frac{b}{a+b}}.\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{a+c}}.\sqrt{\frac{c}{b+c}}\)
Áp dụng bđt Cô-si ngược ta đc
\(A\le\frac{\frac{a}{a+b}+\frac{a}{a+c}}{2}+\frac{\frac{b}{a+b}+\frac{b}{b+c}}{2}+\frac{\frac{c}{c+a}+\frac{c}{b+c}}{2}\)
\(=\frac{\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)+\left(\frac{c}{c+a}+\frac{a}{c+a}\right)}{2}\)
\(=\frac{1+1+1}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\ab+bc+ca=3\end{cases}\Leftrightarrow a=b=c=1}\)
Vậy \(A_{max}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Thôi bt ông giỏi rồi ko cần phải khoe
:))))