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3.
\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)
\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)
\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)
\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)
đặt \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)
\(\frac{x^2}{y-2}+\frac{y^2}{z-2}+\frac{z^2}{x-2};\)áp dụng bdt co-sy \(\frac{x^2}{y-2}+4\left(y-2\right)\ge2.\sqrt{x^2.4}=4x\)
làm tương tự với \(\frac{y^2}{z-2}+4\left(z-2\right)\ge4y;\frac{z^2}{x-2}+4\left(x-2\right)\ge4x\)
=> M +4(x+y+z -6) \(\ge4\left(x+y+z\right)\)<=> M \(\ge24\)
dấu '=' khi x=y=z=4 hay a=b=c = 16
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 <=> \(a=b=c=\sqrt{3}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\frac{5}{4}\left(a+b\right)\)
Tương tự cộng vế theo vế thì
\(M\ge\frac{5}{4}\left(2a+2b+2c\right)=\frac{5}{2}\left(a+b+c\right)=\frac{5}{2}\cdot2019\)
Dấu "=" xảy ra tại \(a=b=c=\frac{2019}{3}\)
bài 4 có trên mạng nha chị.tí e làm cách khác
bài 5 chị tham khảo bđt min cop ski r dùng svác là ra ạ.giờ e coi đá bóng,coi xong nghĩ tiếp ạ.
2. \(BĐT\Leftrightarrow\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\ge1\)
Đặt\(\frac{2}{a}=x;\frac{2}{b}=y;\frac{2}{c}=z\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=8\end{cases}}\)
Ta cần chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge1\Leftrightarrow\left(yz+y+z+1\right)+\left(zx+z+x+1\right)+\left(xy+x+y+1\right)\ge xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\)\(\Leftrightarrow x+y+z\ge6\)(Đúng vì \(x+y+z\ge3\sqrt[3]{xyz}=6\))
Đẳng thức xảy ra khi x = y = z = 2 hay a = b = c = 1
3. Ta có: \(a+b+c\le\sqrt{3}\Rightarrow\left(a+b+c\right)^2\le3\)
Ta có đánh giá quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Từ đó suy ra \(ab+bc+ca\le1\)
\(A=\frac{\sqrt{a^2+1}}{b+c}+\frac{\sqrt{b^2+1}}{c+a}+\frac{\sqrt{c^2+1}}{a+b}\ge\frac{\sqrt{a^2+ab+bc+ca}}{b+c}+\frac{\sqrt{b^2+ab+bc+ca}}{c+a}+\frac{\sqrt{c^2+ab+bc+ca}}{a+b}\)\(=\frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}+\frac{\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}+\frac{\sqrt{\left(c+a\right)\left(c+b\right)}}{a+b}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=3\)Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
- Áp dụng bđt cộng mẫu
Cho \(x_1;x_2;x_3\in R \)
\(\hept{\begin{cases}\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}\ge\frac{\left(x_1+x_2\right)^2}{y_1+y_2}\left(1\right)\\\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}\ge\frac{\left(x_1+x_2+x_3\right)^2}{\left(y_1+y_2+y_3\right)}\left(2\right)\end{cases}}\)
và \(y_1;y_2;y_3\in R\)
CM : +) \(\left(1\right)\Leftrightarrow\left(y_1+y_2\right)\left(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}\right)\ge\left(x_1+x_2\right)^2\)
\(\Leftrightarrow x_1^2+x_2^2+\frac{y_2}{y_1}x_1^2+\frac{y_1}{y_2}x_2^2\ge x_1^2+x_2^2+2x_1x_2\)
\(\Leftrightarrow\frac{y_2}{y_1}x_1^2+\frac{y_1}{y_2}x_2^2\ge2x_1x_2\)( đúng do Cauchy )
+) Để CM (2) , ta áp dụng liên tiếp 2 lần (1)
(1) (2)
\(VT\left(2\right)\ge\frac{\left(x_1+x_1\right)^2}{y_1+y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\)
+) Với cách này ra có thể cm bđt " cộng mẫu " tổng quát sau :
\(\frac{x_1^2}{y_1}+......+\frac{x_1^2}{y_2}\ge\frac{\left(x_1+........+x_n\right)^2}{y_1+...........+y_n}\)
- Áp dụng bđt cộng mẫu , ta có :
\(P=\frac{\sqrt{a}^2}{2\sqrt{b}-5}+\frac{\sqrt{b}^2}{2\sqrt{c}-5}+\frac{\sqrt{c}^2}{2\sqrt{a}-5}\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-15}\ge\frac{S^2}{2S-15}\)
( Trong đó \(S=\sqrt{a}+\sqrt{b}+\sqrt{c}>3\frac{5}{2}=\frac{15}{2}\))
- Đặt U = 2S - 15
+) u > 0
+) \(S=\frac{u+15}{2}\)
\(P\ge\frac{1}{4}.\frac{\left(u+15\right)^2}{u}=\frac{1}{4}\left(u+\frac{15^2}{u}+30\right)\)
\(\ge\frac{1}{4}\left(2\sqrt{u.\frac{15^2}{u}}+30\right)\left(Cauchy\right)\)
\(\ge15\)
Ta có: \(a,b,c>\frac{25}{4}\Rightarrow2\sqrt{a}-5>0,2\sqrt{b}-5>0,2\sqrt{c}-5>0\)
Áp dụng BĐT Cô-si cho 2 số dương ta có:
\(\frac{a}{2\sqrt{b}-5}+2\sqrt{b}-5\ge2\sqrt{a}\) (1)
\(\frac{b}{2\sqrt{c}-5}+2\sqrt{c}-5\ge2\sqrt{b}\) (2)
\(\frac{a}{2\sqrt{a}-5}+2\sqrt{a}-5\ge2\sqrt{c}\) (3)
Cộng vế theo vế của (1), (2), (3) ta có: \(Q\ge5.3=15\)
Dấu '=' xảy ra <=> a=b=c=25 ( TMĐK)
Vậy Min Q =15 <=> a=b=c=25
Đặt \(\left\{{}\begin{matrix}2\sqrt{a}-3=x>0\\2\sqrt{b}-3=y>0\\2\sqrt{c}-3=z>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{\left(x+3\right)^2}{4}\\b=\frac{\left(y+3\right)^2}{4}\\c=\frac{\left(z+3\right)^2}{4}\end{matrix}\right.\)
\(A=\frac{\left(x+3\right)^2}{4y}+\frac{\left(y+3\right)^2}{4z}+\frac{\left(z+3\right)^2}{4x}\)
\(Q\ge\frac{\left(x+y+z+9\right)^2}{4\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2+18\left(x+y+z\right)+81}{4\left(x+y+z\right)}\)
\(4Q\ge x+y+z+\frac{81}{x+y+z}+18\ge2\sqrt{\frac{81\left(x+y+z\right)}{x+y+z}}+18=36\)
\(\Rightarrow Q\ge9\Rightarrow Q_{min}=9\) khi \(x=y=z=3\) hay \(a=b=c=9\)
Uầy, chưa ngủ luôn ạ