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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 <=> \(a=b=c=\sqrt{3}\)
mình đánh nhầm, đề là cho a,b,c là các số thực dương tổng bằng 1
Ta có:
Theo bất đẳng thức Cô - si, ta có: \(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\le\frac{a+b+a+c}{2}+\frac{b+c}{2}=1\)
\(\Rightarrow\sqrt{a}\left(\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{bc}\right)\le\sqrt{a}\)hay \(\sqrt{a^2+abc}+\sqrt{abc}\le\sqrt{a}\)
Tương tự ta có: \(\sqrt{b^2+abc}+\sqrt{abc}\le\sqrt{b}\);\(\sqrt{c^2+abc}+\sqrt{abc}\le\sqrt{c}\)
Mà \(abc\le\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}\Rightarrow\sqrt{abc}\le\frac{1}{3\sqrt{3}}\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\le3\left(a+b+c\right)=3\)\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\le\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)
\(3a^2+4ab+b^2=3a^2+3ab+ab+b^2=3a\left(a+b\right)+b\left(a+b\right)=\left(3a+b\right)\left(a+b\right)\)
xong AM -GM
Ez to prove \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ca\)
\(\Leftrightarrow\frac{6054}{3}\ge ab+bc+ca\Leftrightarrow ab+ca+bc\le2018\)
Khi đó: \(\frac{2a}{\sqrt{a^2+2018}}\le\frac{2a}{\sqrt{a^2+ab+bc+ca}}=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+b}+\frac{a}{a+c}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(P\le\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}=3\)
Vì \(a;b;c>0\Rightarrow2ab\le\frac{\left(a+b\right)^2}{2}\) thay vào \(\sqrt{a^2+4ab+b^2}\)ta có:
\(\sqrt{a^2+4ab+b^2}=\sqrt{\left(a+b\right)^2+2ab}\)
\(\le\sqrt{\left(a+b\right)^2+\frac{\left(a+b\right)^2}{2}}=\sqrt{\frac{3\left(a+b\right)^2}{2}}=\left(a+b\right).\sqrt{\frac{3}{2}}\)
Tương tự: \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}.\left(b+c\right)\)
\(\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}.\left(c+a\right)\)
\(\Rightarrow P\le\sqrt{\frac{3}{2}}.\left(a+b\right)+\sqrt{\frac{3}{2}}.\left(b+c\right)+\sqrt{\frac{3}{2}}.\left(c+a\right)\)
\(\le\sqrt{\frac{3}{2}}.\left(2a+2b+2c\right)=\sqrt{\frac{3}{2}}.6=\sqrt{216}=6\sqrt{6}\)Vì a+b+c=6
Dấu = xảy ra khi a=b=c=2
Vây ......
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
Có \(\sqrt{a^2+4ab+b^2}=\sqrt{\left(\frac{3}{2}a^2+3ab+\frac{3}{2}b^2\right)-\left(\frac{1}{2}a^2-ab+\frac{1}{2}b^2\right)}\)
\(=\sqrt{\frac{3}{2}\left(a+b\right)^2-\frac{1}{2}\left(a-b\right)^2}\le\sqrt{\frac{3}{2}\left(a+b\right)^2}=\sqrt{\frac{3}{2}}\left(a+b\right)\)
Tương tự, ta có : \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}\left(b+c\right);\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}\left(c+a\right)\)
\(\Rightarrow\)\(S\le\sqrt{\frac{3}{2}}\left(a+b\right)+\sqrt{\frac{3}{2}}\left(b+c\right)+\sqrt{\frac{3}{2}}\left(c+a\right)=\sqrt{\frac{3}{2}}.2\left(a+b+c\right)=6\sqrt{6}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)