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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 <=> \(a=b=c=\sqrt{3}\)
\(P=\dfrac{a}{a+\sqrt{2018a+bc}}+\dfrac{b}{b+\sqrt{2018b+ca}}+\dfrac{c}{c+\sqrt{2018c+ab}}\)
\(=\dfrac{a}{a+\sqrt{a.\left(a+b+c\right)+bc}}+\dfrac{b}{b+\sqrt{b.\left(a+b+c\right)+ca}}+\dfrac{c}{c+\sqrt{c.\left(a+b+c\right)+ab}}\)
\(=\dfrac{a}{a+\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{b+\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{c+\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{a\left(\sqrt{a^2+ab+bc+ca}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{b^2+ab+bc+ca}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{c^2+ab+bc+ca}-c\right)}{ab+bc+ca}\)
\(=\dfrac{a\left(\sqrt{\left(a+b\right)\left(a+c\right)}-a\right)}{ab+bc+ca}+\dfrac{b\left(\sqrt{\left(b+c\right)\left(b+a\right)}-b\right)}{ab+bc+ca}+\dfrac{c\left(\sqrt{\left(c+a\right)\left(c+b\right)}-c\right)}{ab+bc+ca}\)
\(\le\dfrac{a\left(\dfrac{2a+b+c}{2}-a\right)}{ab+bc+ca}+\dfrac{b\left(\dfrac{2b+c+a}{2}-b\right)}{ab+bc+ca}+\dfrac{c\left(\dfrac{2c+b+a}{2}-c\right)}{ab+bc+ca}\)
\(=\dfrac{ab+ac}{2\left(ab+bc+ca\right)}+\dfrac{bc+ba}{2\left(ab+bc+ca\right)}+\dfrac{ca+cb}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{2\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=1\)
\(maxP=1\Leftrightarrow a=b=c=\dfrac{2018}{3}\)
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 \(a=b=c=1\)
\(1,\hept{\begin{cases}10x^2+5y^2-2xy-38x-6y+41=0\left(1\right)\\3x^2-2y^2+5xy-17x-6y+20=0\left(2\right)\end{cases}}\)
Giải (1) : \(10x^2+5y^2-2xy-38x-6y+41=0\)
\(\Leftrightarrow10x^2-2x\left(y+19\right)+5y^2-6y+41=0\)
Coi pt trên là pt bậc 2 ẩn x
Có \(\Delta'=\left(y+19\right)^2-50y^2+60y-410\)
\(=-49y^2+98y-49\)
\(=-49\left(y-1\right)^2\)
pt có nghiệm \(\Leftrightarrow\Delta'\ge0\)
\(\Leftrightarrow-49\left(y-1\right)^2\ge0\)
\(\Leftrightarrow y=1\)
Thế vào pt (2) được x = 2
\(2,\)Đặt\(\left(a\sqrt{a};b\sqrt{b};c\sqrt{c}\right)\rightarrow\left(x;y;z\right)\left(x,y,z>0\right)\)
\(\Rightarrow xy+yz+zx=1\)
Khi đó \(P=\frac{x^4}{x^2+y^2}+\frac{y^4}{y^2+z^2}+\frac{z^4}{x^2+z^2}\)
Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(x;y;z>0\right)\left(Cauchy-engel-type_3\right)\)được
\(P\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{2}\)
Áp dụng bđt x2 + y2 + z2 > xy + yz + zx (tự chứng minh) ta được
\(P\ge\frac{x^2+y^2+z^2}{2}\ge\frac{xy+yz+zx}{2}=\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}xy+yz+zx=1\\x=y=z\end{cases}}\)
\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\sqrt{a^3}=\sqrt{b^3}=\sqrt{c^3}=\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow a^3=b^3=c^3=\frac{1}{3}\)
\(\Leftrightarrow a=b=c=\frac{1}{\sqrt[3]{3}}\)
Vậy \(P_{min}=\frac{1}{2}\Leftrightarrow a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(A=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}+\frac{c}{\sqrt{3+c^2}}\)
\(=\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+bc+ca+ab}}+\frac{c}{\sqrt{c^2+ca+ab+bc}}\)
\(=\frac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\sqrt{b}\cdot\sqrt{b}}{\sqrt{\left(b+c\right)\left(a+b\right)}}+\frac{\sqrt{c}\cdot\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(=\frac{\sqrt{a}}{\sqrt{a+b}}\cdot\frac{\sqrt{a}}{\sqrt{c+a}}+\frac{\sqrt{b}}{\sqrt{b+c}}\cdot\frac{\sqrt{b}}{\sqrt{a+b}}+\frac{\sqrt{c}}{\sqrt{c+a}}\cdot\frac{\sqrt{c}}{\sqrt{c+b}}\)
\(\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}+\frac{\frac{b}{b+c}+\frac{b}{a+b}}{2}+\frac{\frac{c}{c+a}+\frac{c}{b+c}}{2}\)
\(=\frac{\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}}{2}=\frac{3}{2}\)
Vậy Max A = 3/2 khi a = b = c = 1. (Max not Min)
Cho mk k nhé!
4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13
1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)
Có: \(\frac{2018a+3}{1+b^2}=2018a+3-\frac{b^2\left(2018a+3\right)}{1+b^2}\) (Làm tắt ráng hiểu ^^)
\(\ge2018a+3-\frac{b^2\left(2018a+3\right)}{2b}\left(Cauchy\right)\)
\(=2018a+3-\frac{b\left(2018a+3\right)}{2}\)
\(=2018a+3-\frac{2018ab+3b}{2}\)
Tương tự \(\frac{2018b+3}{1+c^2}\ge2018b+3-\frac{2018bc+3b}{2}\)
\(\frac{2018c+3}{1+a^2}\ge2018c+3-\frac{2018ac+3a}{2}\)
CỘng vế với vế của các bđt trên lại ta được
\(A\ge2018\left(a+b+c\right)+9-\frac{2018\left(ab+bc+ca\right)+3\left(a+b+c\right)}{2}\)
\(=2018\left(a+b+c\right)+9-\frac{6054+3\left(a+b+c\right)}{2}\)
\(=2018\left(a+b+c\right)-\frac{3\left(a+b+c\right)}{2}-3018\)
\(=\frac{4033\left(a+b+c\right)}{2}-3018\)
Ta có bđt phụ : \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\)(1)
Thật vậy \(\left(1\right)\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(Luôn đúng)
Nên (1) được chứng minh
ÁP dụng (1) ta được \(A\ge\frac{4033\left(a+b+c\right)}{2}-3018\ge\frac{4033}{2}\sqrt{3\left(ab+bc+ca\right)}-3018\)
\(=\frac{4033}{2}\sqrt{3.3}-3018\)
\(=\frac{6063}{2}\)
Dấu "='' xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b=c\\ab+bc+ca=3\end{cases}\Leftrightarrow}a=b=c=1\)
Vậy \(A_{min}=\frac{6063}{2}\Leftrightarrow a=b=c=1\)