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vì \(c\le a\)nên \(\frac{1}{\left(c+1\right)^2}\ge\frac{1}{\left(a+1\right)^2}\)
\(VT\ge\frac{2}{\left(a+1\right)^2}+\frac{2}{\left(b+1\right)^2}+\frac{2}{\left(c+1\right)^2}\)
Áp dụng BĐT AM-GM: \(\frac{1}{\left(a+1\right)^2}+\frac{1}{\left(b+1\right)^2}+\frac{1}{\left(c+1\right)^2}\ge\frac{1}{\left(a+1\right)\left(b+1\right)}+\frac{1}{\left(b+1\right)\left(c+1\right)}+\frac{1}{\left(c+1\right)\left(a+1\right)}\)
\(=\frac{a+b+c+3}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{a+b+c+3}{abc+a+b+c+4}\)(*)
Từ giả thiết: ab+bc+ca=3.Áp dụng BĐT AM-GM:\(3=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow abc\le1\)
và có BĐT \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=9\)\(\Leftrightarrow a+b+c\ge3\)
\(\Rightarrow a+b+c\ge3\ge3abc\)
từ (*): \(\frac{a+b+c+3}{abc+a+b+c+4}\ge\frac{a+b+c+3}{\frac{a+b+c}{3}+a+b+c+4}=\frac{3\left(a+b+c+3\right)}{4\left(a+b+c\right)+12}=\frac{3}{4}\)
do đó \(VT\ge2.\frac{3}{4}=\frac{3}{2}\)
Dấu = xảy ra khi a=b=c=1
nguồn: Hữu Đạt
Áp dụng BĐT AM-GM (Cô si): \(A\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
\(=3\sqrt[3]{\frac{1}{a\left(b+c\right).b\left(c+a\right).c\left(a+b\right)}}=\frac{3}{\sqrt[3]{\left(ab+ca\right)\left(bc+ab\right)\left(ca+bc\right)}}\)
\(\ge\frac{9}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
P/s: Check giúp em xem có ngược dấu không:v
Cach khac
Dat \(\left(ab;bc;ca\right)\rightarrow\left(x;y;z\right)\)
\(\Rightarrow\hept{\begin{cases}x+y+z=3\\x^2+y^2+z^2\ge3\\xyz\le1\end{cases}}\)
Ta co:
\(A=\frac{1}{ab+b^2}+\frac{1}{bc+c^2}+\frac{1}{ca+a^2}\)
\(=\frac{1}{x+\frac{xy}{z}}+\frac{1}{y+\frac{yz}{x}}+\frac{1}{z+\frac{zx}{y}}\ge\frac{9}{3+xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{3+3}=\frac{3}{2}\)
Dau '=' xay ra khi \(a=b=c=1\)
Vay \(A_{min}=\frac{3}{2}\)khi \(a=b=c=1\)
Đặt \(\frac{1}{a}=x>0;\frac{1}{b}=y>0;\frac{1}{c}=z>0\)
Từ giả thiết ta có: \(7\left(x^2+y^2+z^2\right)=6\left(xy+yz+zx\right)+2015\le6\left(x^2+y^2+z^2\right)+2015\)
\(\Leftrightarrow x^2+y^2+z^2\le2015\)
Ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}=\frac{1}{\sqrt{\left(4a^2+b^2\right)+\left(2a^2+2b^2\right)}}\le\frac{1}{\sqrt{4a^2+b^2+4ab}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)=\frac{1}{9}\left(2x+y\right)\)
Tương tự thì: \(\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\frac{1}{9}\left(2y+z\right)\) và \(\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{1}{9}\left(2z+x\right)\)
Cộng từng vế 3 BĐT trên ta có: \(\frac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\frac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\frac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\frac{x+y+z}{3}\le\frac{\sqrt{3\left(x^2+y^2+z^2\right)}}{3}\le\sqrt{\frac{2015}{3}}\)
Vậy max \(P=\sqrt{\frac{2015}{3}}\) , đạt được khi \(a=b=c=\sqrt{\frac{3}{2015}}\)
Dat \(\left(\frac{a}{b};\frac{b}{c};\frac{c}{a}\right)=\left(x;y;z\right)\)
\(\Rightarrow xyz=1\)
\(\Sigma_{cyc}\frac{1}{\frac{a}{b}+\frac{c}{a}+1}=\Sigma_{cyc}\frac{1}{x+y+1}\)
We need to prove:
\(\Sigma_{cyc}\frac{1}{x+y+1}\le1\)
\(\Leftrightarrow\Sigma_{cyc}\frac{x+y}{x+y+1}\ge2\left(M\right)\)
We have:
\(VT_M\ge\frac{\left(\Sigma_{cyc}\sqrt{x+y}\right)^2}{2\Sigma_{cyc}x+3}\)
Now we need to prove
\(\frac{\left(\Sigma_{cyc}\sqrt{x+y}\right)^2}{2\Sigma_{cyc}x+3}\ge2\)
\(\Leftrightarrow\Sigma_{cyc}\sqrt{\left(x+y\right)\left(y+z\right)}\ge\Sigma_{cyc}x+3\left(M_1\right)\)
Consider:
\(VT_{M_1}=\sqrt{\left(x+y\right)\left(y+z\right)}\ge x+y+z+xy+yz+zx\)
Now we need to prove:
\(x+y+z+xy+yz+zx\ge x+y+z+3\)
\(xy+yz+zx\ge3\) (Not fail with xyz=1)
Dau '=' xay ra khi \(\hept{\begin{cases}a=b=c=1\\x=y=z=1\end{cases}}\)
ab+bc+ca=3abc <=> ab+bc+ca-3abc=0 <=> ab-abc+bc-abc+ca-abc=0 <=> ab(1-c)+bc(1-a)+ca(1-b)=0
Vì a,b,c dương => \(\hept{\begin{cases}1-c=0< =>c=1\\1-a=0< =>a=1\\1-b=0< =>b=1\end{cases}}\)
Thay a,b,c vừa tìm được vào biểu thức P <=> P=3/2
áp dụng BDT cô si ta có
\(a^2+1>=2a\)
\(b^2+1>=2b\)
\(c^2+1>=2c\)
do đó P<=\(\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\)
=\(\frac{1}{2}.\frac{3abc}{abc}=1,5\)
dấu = xảy ra khi và chỉ khi a=b=c=1
\(a,M=6+\frac{2x}{x^2+1}\)
ÁP dụng bđt AM-GM ta có
\(M\le6+\frac{2x}{2x}=7\)
Dấu "=" xảy ra khi x=1
b,\(A=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)(bđt Cauchy-Schwarz)
mà \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow A\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi a=b=c