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Đề như này đúng ko \(3\le\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}< 3+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
Dấu \("\ge"\) thứ 2 dấu "=" ko xảy ra
Đặt \(A=\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\)
\(A\ge3\sqrt[3]{\frac{\left(1+\sqrt{a}\right)\left(1+\sqrt{b}\right)\left(1+\sqrt{c}\right)}{\left(1+\sqrt{b}\right)\left(1+\sqrt{c}\right)\left(1+\sqrt{a}\right)}}=3\) \(\left(1\right)\)
CM : \(\frac{1+\sqrt{x}}{1+\sqrt{y}}< 1+\sqrt{x}\) ( với a, b nguyên dương )
\(\Leftrightarrow\)\(\left(1+\sqrt{x}\right)\left(1+\sqrt{y}\right)-\left(1+\sqrt{x}\right)>0\)
\(\Leftrightarrow\)\(\left(1+\sqrt{x}\right)\sqrt{y}>0\) ( luôn đúng với mọi a, b nguyên dương )
\(\Rightarrow\)\(A< 1+\sqrt{a}+1+\sqrt{b}+1+\sqrt{c}=3+\sqrt{a}+\sqrt{b}+\sqrt{c}\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(3\le\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}< 3+\sqrt{a}+\sqrt{b}+\sqrt{c}\) ( đpcm )
Chúc bạn học tốt ~
1) c/m \(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
áp dụng BĐT cô shi cho 2 số thực dương ta có:
\(a+b\ge2\sqrt{ab}\);\(b+c\ge2\sqrt{bc}\);\(a+c\ge2\sqrt{ac}\)
cộng vế vs vế:\(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
↔\(a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
dấu = xảy ra khi a=b=c
vậy...
b)ta có:
\(\frac{1}{\sqrt{1}}>\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>...>\frac{1}{\sqrt{25}}\)→\(A>\frac{1}{\sqrt{25}}+\frac{1}{\sqrt{25}}+...+\frac{1}{\sqrt{25}}\)(25 số hạng)
\(A>\frac{25}{\sqrt{25}}=\sqrt{25}=5\)
vậy.....
1,
\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)
\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)
Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)
Ta có \(a+b+c\le\sqrt{3}\)
\(\Rightarrow\left(a+b+c\right)^2\le3\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\le1\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}\ge ab+bc+ac\)
\(\Rightarrow1\ge ab+bc+ac\)
\(\Rightarrow\left\{\begin{matrix}1+a^2\ge a^2+ab+bc+ac\\1+b^2\ge b^2+ab+bc+ac\\1+c^2\ge c^2+ab+bc+ac\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\sqrt{1+a^2}\ge\sqrt{a^2+ab+bc+ca}\\\sqrt{1+b^2}\ge\sqrt{b^2+ab+bc+ca}\\\sqrt{1+c^2}\ge\sqrt{c^2+ab+bc+ca}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{1+a^2}}\le\frac{a}{\sqrt{a^2+ab+bc+ac}}\\\frac{b}{\sqrt{1+b^2}}\le\frac{b}{\sqrt{b^2+ab+bc+ac}}\\\frac{c}{\sqrt{1+c^2}}\le\frac{c}{\sqrt{c^2+ab+bc+ac}}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{a^2+ab+bc+ca}}+\frac{b}{\sqrt{b^2+ab+bc+ca}}+\frac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{a\left(a+b\right)+c\left(a+b\right)}}+\frac{b}{\sqrt{b\left(b+a\right)+c\left(a+b\right)}}+\frac{c}{\sqrt{c\left(c+a\right)+b\left(c+a\right)}}\)
\(\Rightarrow\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Xét \(\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng bất đẳng thức Cauchy ngược dấu cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\sqrt{\left(a+b\right)\left(a+c\right)}\ge\frac{2a+b+c}{2}\\\sqrt{\left(a+b\right)\left(b+c\right)}\ge\frac{a+2b+c}{2}\\\sqrt{\left(c+a\right)\left(c+b\right)}\ge\frac{a+b+2c}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{2a}{2b+b+c}\\\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\frac{2b}{a+2b+c}\\\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{2c}{a+b+2c}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\)
Chứng minh rằng: \(2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\le\frac{3}{4}\)
Áp dụng bất đẳng thức \(\frac{1}{a+b}\ge\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\frac{a}{2a+b+c}=\frac{a}{a+c+a+b}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\Rightarrow\frac{b}{a+2b+c}=\frac{b}{a+b+b+c}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\)
\(\Rightarrow\frac{c}{a+b+2c}=\frac{c}{a+c+b+c}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{b}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\left(đpcm\right)\)
\(\Rightarrow2\left(\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\right)\le\frac{3}{2}\)
\(\Rightarrow\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\frac{3}{2}\)
Vậy \(\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\le\frac{3}{2}\left(đpcm\right)\)
Lời giải khác:
Áp dụng bđt Cauchy-Schwarz:
\((a^2+1)(1+3)\geq (a+\sqrt{3})^2\)\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{2a}{a+\sqrt{3}}\)
Thực hiện tương tự với các phân thức còn lại:
\(\Rightarrow \frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}}\leq 2\left ( \frac{a}{a+\sqrt{3}}+\frac{b}{b+\sqrt{3}}+\frac{c}{c+\sqrt{3}} \right )=2A\) $(1)$
Lại có:
\(\)\(A=\left ( 1-\frac{\sqrt{3}}{a+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{b+\sqrt{3}} \right )+\left ( 1-\frac{\sqrt{3}}{c+\sqrt{3}} \right )=3-\sqrt{3}\left ( \frac{1}{a+\sqrt{3}}+\frac{1}{b+\sqrt{3}}+\frac{1}{c+\sqrt{3}} \right )\)
Cauchy-Schwarz kết hợp với \(a+b+c\leq \sqrt{3}\):
\(A\leq 3-\frac{9\sqrt{3}}{a+b+c+3\sqrt{3}}\leq 3-\frac{9\sqrt{3}}{4\sqrt{3}}=\frac{3}{4}\) $(2)$
Từ \((1),(2)\Rightarrow \text{VT}\leq 2A\leq \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
bài này hay đấy
Áp dụng BĐT Cô-si cho 3 số không âm, ta có :
\(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\ge3\sqrt[3]{\frac{1+\sqrt{a}}{1+\sqrt{b}}.\frac{1+\sqrt{b}}{1+\sqrt{c}}.\frac{1+\sqrt{c}}{1+\sqrt{a}}}=3\)
Chứng minh \(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\le3+a+b+c\)( 1 )
đặt \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)( x,y,z \(\ge\)0 )
do a,b,c là số nguyên
Nếu a = b = c = 0 thì x = y = z = 0 nên ( 1 ) đúng
Nếu a,b,c không đồng thời bằng 0 \(\Rightarrow\)x+ y + z \(\ge\)1
Ta có : VT ( 1 )
\(\Leftrightarrow\frac{\left(1+x\right)\left(1+y\right)-\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)\left(1+z\right)-\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)\left(1+x\right)-\left(1+z\right)x}{1+z}\)
\(=3+x+y+z-\left[\frac{\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)x}{1+x}\right]\)
\(\le3+x+y+z-\frac{\left(1+x\right)y+\left(1+y\right)z+\left(1+z\right)x}{1+x+y+z}=3+x+y+z-\frac{x+y+z+xy+yz+xz}{1+x+y+z}\)
\(=3+\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le3+x^2+y^2+z^2\)
Cần chứng minh : \(\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le x^2+y^2+z^2\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)
Mà \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge1.\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)
suy ra đpcm