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Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiếp lập 2 BĐT còn lại:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+c}+\dfrac{b}{a+b}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Ta có: \(\frac{1}{m}+\frac{1}{n}=\frac{1}{2}\)
\(\Rightarrow\frac{m+n}{mn}=\frac{1}{2}\)
\(\Leftrightarrow mn=2\left(m+n\right)\)
\(\Rightarrow2mn=4\left(m+n\right)\)
Từ Phương trình 1 lập \(\Delta_1\)
\(\Delta_1=m^2-4n\)
Phương trình 2 có \(\Delta_2=n^2-4m\)
lấy \(\Delta_1+\Delta_2\)
\(=m^2+n^2-4m-4n\)
\(=m^2-4\left(m+n\right)+n^2\)
\(=m^2-2mn+n^2\)
\(=\left(m-n\right)^2\ge0\)
vậy tồn tại delta1 hoặc delta 2 dương nên một trong 2 phương trình đã cho có ít nhất 1 phương trình có nghiệm
Ta sẽ chứng minh :
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) với x, y > 0
Thật vậy : \(x+y+z\ge3\sqrt[3]{xyz}\)( bđt Cô - si )
Và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{abc}}\) ( bđt Cô - si )
\(\Rightarrow x+y+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\) ( Dấu " = " \(\Leftrightarrow x=y=z\) )
Ta có :
\(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
( Dấu " = " xay ra khi a=b)
Tương tự ta cũng có :
\(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\) ( Dấu " = " xảy ra khi b=c)
\(\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\) ( Dấu " = " xay ra khi c = a )
\(VT=\sum_{cyc}\frac{1}{\sqrt{5a^2+2ab+b^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu " = " xay ra khi \(a=b=c=\frac{2}{3}\)
Chúc bạn học tốt !!
\(\frac{1}{\sqrt{4a^2+2ab+b^2+a^2+b^2}}\le\frac{1}{\sqrt{4a^2+2ab+b^2+2ab}}=\frac{1}{\sqrt{\left(2a+b\right)^2}}=\frac{1}{2a+b}=\frac{1}{a+a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
\(\Rightarrow VT\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}+\frac{2}{b}+\frac{1}{c}+\frac{2}{c}+\frac{1}{a}\right)=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{2}{3}\)
Từ giả thiết suy ra : \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Nên ta có : \(\frac{\sqrt{1+x^2}}{x}=\sqrt{\frac{1}{x^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}\le\frac{1}{2}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow y=z\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}\le\frac{1}{2}\left(\frac{4}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có :
\(\frac{1+\sqrt{1+y^2}}{y}\le\frac{1}{2}\left(\frac{1}{x}+\frac{4}{y}+\frac{1}{z}\right);\frac{1+\sqrt{1+z^2}}{z}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}\right)\)
Vậy ta có :
\(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Ta có :
\(\left(x+y+z\right)^2-3\left(xy+yz+xx\right)=...=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\ge0\)
Nên \(\left(x+y+x\right)^2\ge3\left(xy+yz+xx\right)\)
\(\Rightarrow\left(xyz\right)^2\ge3\left(xy+yz+xz\right)\Rightarrow3\frac{xy+yz+xz}{xyz}\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le xyz\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le xyz\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Chúc bạn học tốt !!
\(\frac{1+\frac{1}{2}.2.\sqrt{1+x^2}}{x}\le\frac{1+\frac{1}{4}\left(x^2+5\right)}{x}=\frac{x}{4}+\frac{9}{4x}\)
\(\Rightarrow VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4xyz}=\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4\left(x+y+z\right)}\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{3\left(x+y+z\right)^2}{4\left(x+y+z\right)}=x+y+z=xyz\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
+ \(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2\left(ab+b+1\right)}\) . Dấu "=" \(\Leftrightarrow a=b=1\)
+ Tương tự : \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\). Dấu "=" \(\Leftrightarrow b=c=1\)
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ca+a+1\right)}\). Dấu "=" \(c=a=1\)
Do đó : \(VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{abc\cdot b+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)=\frac{1}{2}\)
Dấu "=" \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT BCS dạng phân thức ta được:
\(\frac{1}{4a^2+b^2+c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2a^2+\left(a^2+b^2\right)+\left(c^2+a^2\right)}\le\frac{1}{9}\left(\frac{a^2}{2a^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+a^2}\right)\)
\(\frac{1}{a^2+4b^2+c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2b^2+\left(a^2+b^2\right)+\left(c^2+b^2\right)}\le\frac{1}{9}\left(\frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{c^2+b^2}\right)\)
\(\frac{1}{a^2+b^2+4c^2}=\frac{1}{9}.\frac{\left(a+b+c\right)^2}{2c^2+\left(a^2+c^2\right)+\left(c^2+b^2\right)}\le\frac{1}{9}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{c^2+b^2}+\frac{c^2}{2c^2}\right)\)
Cộng các BĐT trên theo vế ta được:
\(\frac{1}{4a^2+b^2+c^2}+\frac{1}{a^2+4b^2+c^2}+\frac{1}{a^2+b^2+4c^2}\le\frac{1}{2}\)
Dấu $"="$ xảy ra \(\Leftrightarrow a=b=c\)
Băng :v
Ta có
\(\frac{1+m^2}{1+n^2}=1+m^2-\frac{n^2\left(1+m^2\right)}{1+n^2}\le1+m^2-\frac{n^2\left(1+m^2\right)}{2}\)
Tương tự ta có
\(\frac{1+n^2}{1+p^2}\le1+n^2-\frac{p^2\left(1+n^2\right)}{2}\)
\(\frac{1+p^2}{1+m^2}\le1+p^2-\frac{m^2\left(1+p^2\right)}{2}\)
\(\Rightarrow A\le3+m^2+n^2+p^2-\frac{n^2\left(1+m^2\right)+p^2\left(1+n^2\right)+m^2\left(1+p^2\right)}{2}\)
\(=\frac{m^2+n^2+p^2-\left(m^2N^2+n^2p^2+p^2m^2\right)}{2}+3\)
\(\le\frac{m^2+n^2+p^2+2\left(mn+np+pm\right)}{2}+3\)
\(=\frac{\left(m+n+p\right)^2}{2}+3=\frac{1}{2}+3=\frac{7}{2}\)
\(a,b,c\in\left[0,1\right]\) do đó \(a^2+b^2+c^2\le a+b+c=1\)
Ta có: \(T=\text{∑}\left(a^2+1-\frac{b^2a^2+b^2}{1+b^2}\right)\)\(\le\text{∑}a^2+3-\text{∑}\frac{b^2a^2+b^2}{2}\)
\(=3+\frac{\text{∑}a^2-\text{∑}a^2b^2}{2}\le3+\frac{1}{2}\le\frac{7}{2}\)