cho 3 số thức a,b,>0 cm BĐt sau \(\frac{1}{1+a^2}+\frac{1}{1+b^2}>=\frac{2}{1+ab}\) khi a.b>=1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
a/ \(VT=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow VT\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{3}{4}\)
b/ \(VT\le\frac{ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{ca}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(VT\le\frac{a}{4}+\frac{b}{4}+\frac{b}{4}+\frac{c}{4}+\frac{c}{4}+\frac{a}{4}=\frac{a+b+c}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
sử dụng hệ quả bun-nhi-a ta có:
VT\(\ge\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)+\left(ab+bc+ca\right)}\)
mà từ giả thiết , kết hợp với bất đẳng thức , ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)=>\(a+b+c\ge9\)
mặt khác: ab+bc+ca\(\le\frac{\left(a+b+c\right)^2}{3}\)
=> VT\(\ge\)\(\frac{3\left(a+b+c\right)^2}{\left(a+b+c\right)\left(a+b+c+3\right)}\ge\frac{3\left(a+b+c\right)^2}{\left(a+b+c\right)\frac{4\left(a+b+c\right)}{3}}=\frac{a+b+c}{4}\)(dpcm)
Đây nhé
Đặt b + c = x ; c + a = y ; a + b = z
\(\Rightarrow\hept{\begin{cases}x+y=2c+b+a=2c+z\\y+z=2a+b+c=2a+x\\x+z=2b+a+c=2b+y\end{cases}}\)
\(\Rightarrow\frac{x+y-z}{2}=c;\frac{y+z-x}{2}=a;\frac{x+z-y}{2}=b\)
Thay vào PT đã cho ở đề bài , ta có :
\(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\)
\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
( cái này cô - si cho x/y + /x ; x/z + z/x ; y/z + z/y)
d/ Đặt \(x=a+b\) , \(y=b+c\) , \(z=c+a\)
thì : \(a=\frac{x+z-y}{2}\) ; \(b=\frac{x+y-z}{2}\) ; \(c=\frac{y+z-x}{2}\)
Ta có : \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}+\frac{\frac{y+z-x}{2}}{x}\)
\(=\frac{z+x-y}{2y}+\frac{x+y-z}{2z}+\frac{y+z-x}{2x}=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{z}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}-3\right)\)
\(=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}\right)-\frac{3}{2}\ge\frac{1}{2}.6-\frac{3}{2}=\frac{3}{2}\)
b/ \(a^2\left(1+b^2\right)+b^2\left(1+c^2\right)+c^2\left(1+a^2\right)\ge6abc\)
\(\Leftrightarrow\left(a^2b^2-2abc+c^2\right)+\left(b^2c^2-2abc+a^2\right)+\left(c^2a^2-2abc+b^2\right)\ge0\)
\(\Leftrightarrow\left(ab-c\right)^2+\left(bc-a\right)^2+\left(ca-b\right)^2\ge0\) (luôn đúng)
Vậy bđt ban đầu dc chứng minh.
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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
Ta có:
\(\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\le1\)
\(\Leftrightarrow\frac{2}{a^2+2}+\frac{2}{b^2+2}+\frac{2}{c^2+2}\le2\)
\(\Leftrightarrow1-\frac{2}{a^2+2}+1-\frac{2}{b^2+2}+1-\frac{2}{c^2+2}\ge1\)
\(\Leftrightarrow\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge1\)
Ta cần cm bđt trên đúng.Thật vậy
\(\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+6}=\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=1\)
\("="\Leftrightarrow a=b=c=1\)
a. Ta có: \(A=\sqrt{\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{2006^2}\right)}=\sqrt{\frac{1}{2}.\frac{3}{2}.\frac{2}{3}.\frac{4}{3}...\frac{2015}{2016}.\frac{2017}{2016}}\)
\(=\sqrt{\frac{1}{2}.\frac{2017}{2016}}=\sqrt{\frac{2017}{4032}}\)
b. Với b > 0 thì a > 0, ta có: \(B=\frac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{b}-\frac{\sqrt{a}}{\sqrt{b}}=\frac{-\sqrt{b}}{\sqrt{b}}=-1\)
Với b < 0 thì a < 0, ta có: \(B=\frac{\sqrt{ab}-\sqrt{b^2}}{b}-\frac{\sqrt{ab}}{\sqrt{b^2}}=\frac{\sqrt{ab}-\sqrt{b^2}}{b}+\frac{\sqrt{ab}}{b}=\frac{2\sqrt{ab}+b}{b}\)
1. BĐT ban đầu
<=> \(\left(\frac{1}{3}-\frac{b}{a+3b}\right)+\left(\frac{1}{3}-\frac{c}{b+3c}\right)+\left(\frac{1}{3}-\frac{a}{c+3a}\right)\ge\frac{1}{4}\)
<=>\(\frac{a}{a+3b}+\frac{b}{b+3c}+\frac{c}{c+3a}\ge\frac{3}{4}\)
<=> \(\frac{a^2}{a^2+3ab}+\frac{b^2}{b^2+3bc}+\frac{c^2}{c^2+3ac}\ge\frac{3}{4}\)
Áp dụng BĐT buniacoxki dang phân thức
=> BĐT cần CM
<=> \(\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3\left(ab+bc+ac\right)}\ge\frac{3}{4}\)
<=> \(a^2+b^2+c^2\ge ab+bc+ac\)luôn đúng
=> BĐT được CM
2) \(a+b+c\le ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(\left(a+b+c\right)^2-3\left(a+b+c\right)\ge0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left(a+b+c-3\right)\ge0\)\(\Leftrightarrow\)\(a+b+c\ge3\)
ko mất tính tổng quát giả sử \(a\ge b\ge c\)
Có: \(3\le a+b+c\le ab+bc+ca\le3a^2\)\(\Leftrightarrow\)\(3a^2\ge3\)\(\Leftrightarrow\)\(a\ge1\)
=> \(\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{3}{1+2a}\le1\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=1\)
\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)
\(\Leftrightarrow\frac{\left(1+b^2\right)\left(1+ab\right)+\left(1+a^2\right)\left(1+ab\right)-2\left(1+a^2\right)\left(1+b^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)
\(\Leftrightarrow ab^3+b^2+ab+1+a^3b+a^2+ab+1-2a^2b^2-2a^2-2b^2-2\ge0\)
\(\Leftrightarrow a^3b+ab^3-2a^2b^2-a^2-b^2+2ab\ge0\)
\(\Leftrightarrow ab\left(a-b\right)^2-\left(a-b\right)^2\ge0\)
\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)đúng do \(ab\ge1,\left(a-b\right)^2\ge0\).
Do biến đổi tương đương, bất đẳng thức cuối đúng nên bất đẳng thức cần chứng minh cũng đúng.
Ta có đpcm.