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\(gt\Rightarrow1=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{1}{a^2}+1=\frac{1}{a^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\)
\(\frac{1}{ab}\sqrt{\frac{\left(a^2+1\right)\left(b^2+1\right)}{c^2+1}}=\sqrt{\frac{\left(1+\frac{1}{a^2}\right)\left(1+\frac{1}{b^2}\right)}{c^2\left(1+\frac{1}{c^2}\right)}}\)
\(=\frac{1}{c}.\sqrt{\frac{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{a}\right)\left(\frac{1}{b}+\frac{1}{c}\right)}{\left(\frac{1}{c}+\frac{1}{a}\right)\left(\frac{1}{c}+\frac{1}{b}\right)}}=\frac{1}{c}\sqrt{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{bc}+\frac{1}{ca}\)
Tương tự với các cụm còn lại, ta được
\(A=2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)
bài này khó thật, nhưng bạn đừng buồn, sẽ có nhiều bạn khác giúp bạn
nha Nguyễn Quang Linh à
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
\(\frac{1}{a-1}=\left(1-\frac{1}{b-1}\right)+\left(1-\frac{1}{c-1}\right)=\frac{b-2}{b-1}+\frac{c-2}{c-1}\ge2\sqrt{\frac{\left(b-2\right)\left(c-2\right)}{\left(b-1\right)\left(c-1\right)}}\)
Tương tự với \(\frac{1}{b-1};\text{ }\frac{1}{c-1}\)
Rồi nhân theo vế 3 bất đẳng thức:
\(\frac{1}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\ge8\sqrt{\frac{\left(a-2\right)^2\left(b-2\right)^2\left(c-2\right)^2}{\left(a-1\right)^2\left(b-1\right)^2\left(c-1\right)^2}}=8\frac{\left(a-2\right)\left(b-2\right)\left(c-2\right)}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}\)
\(\Rightarrow\left(a-2\right)\left(b-2\right)\left(c-2\right)\le\frac{1}{8}\)
Vậy GTLN của H là 0,125.
Đẳng thức xảy ra khi \(a=b=c=\frac{5}{2}.\)
Áp dụng Bunhia cho bộ số (1;1;1) vfa (a;b;c) ta có 3(a2+b2+c2) >= (a+b+c)2
=> 3(2a2+b2) >=(2a+b2); 3(2b2+c2) >= (2b+c)2; 3(2c2+a2) >= (2c+a)2
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)
Ta có \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{1}{x+y+z}\)
=> \(P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+b}\le\frac{1}{9}\left[\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\right]\)
=> \(P\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(I\right)\)
Ta có \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
\(=3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\left(II\right)\)
Áp dụng Bunhia cho bộ số (1;1;1) và \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)
Ta được \(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)\(\Rightarrow\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
=> \(10\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\left(III\right)\)
Từ (I)(II)(III) => \(3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+2015\ge10\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\cdot2015\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\sqrt{3\cdot2015}\left(IV\right)\)
Từ (I)(IV) => \(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}\cdot\sqrt{3\cdot2015}=\sqrt{\frac{2015}{3}}\)
Vậy GTNN của P=\(\sqrt{\frac{2015}{3}}\)khi a=b=c và \(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+2015\)
=> \(a=b=c=\sqrt{\frac{3}{2015}}\)
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}=a\sqrt{\frac{1}{a+b}.\frac{1}{c+a}}\le\frac{\frac{a}{a+b}+\frac{a}{c+a}}{2}\)
Tương tự 2 cái còn lại cộng lại ta đc \(VT\le\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
Cach khac
Dat \(P=\frac{a}{\sqrt{bc\left(1+a^2\right)}}+\frac{b}{\sqrt{ca\left(1+b^2\right)}}+\frac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Ta co:
\(a+b+c=abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Dat \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\)
\(\Rightarrow xy+yz+zx=1\)
\(\Rightarrow P=\sqrt{\frac{yz}{1+x^2}}+\sqrt{\frac{zx}{1+y^2}}+\sqrt{\frac{xy}{1+z^2}}\)
Ta lai co:
\(\sqrt{\frac{yz}{1+x^2}}=\sqrt{\frac{yz}{xy+yz+zx+x^2}}=\sqrt{\frac{yz}{\left(x+y\right)\left(z+x\right)}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{z}{z+x}\right)\)
Tuong tu:
\(\sqrt{\frac{zx}{1+y^2}}\le\frac{1}{2}\left(\frac{z}{y+z}+\frac{x}{x+y}\right)\)
\(\sqrt{\frac{xy}{1+z^2}}\le\frac{1}{2}\left(\frac{x}{z+x}+\frac{y}{y+z}\right)\)
\(\Rightarrow P\le\frac{1}{2}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{2}\)
Dau '=' xay ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(\Rightarrow a=b=c=\sqrt{3}\)
Vay \(P_{min}=\frac{3}{2}\)khi \(a=b=c=\sqrt{3}\)
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 <=> \(a=b=c=\sqrt{3}\)
\(\frac{1}{\left(a+1\right)^2+b^2+1}+\frac{1}{\left(b+1\right)^2+c^2+1}+\frac{1}{\left(c+1\right)^2+a^2+1}\)
\(=\frac{1}{a^2+b^2+2a+2}+\frac{1}{b^2+c^2+2b+2}+\frac{1}{c^2+a^2+2c+2}\)
\(\le\frac{1}{2ab+2a+2}+\frac{1}{2bc+2b+2}+\frac{1}{2ac+2c+2}\)
\(=\frac{1}{2}\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\right)=\frac{1}{2}\)
\("="\Leftrightarrow a=b=c=1\)