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Cho a,b,c>0 t/m a+b+c=3
CMR: \(\dfrac{a+1}{b^2+1}\)+\(\dfrac{b+1}{c^2+1}\)+\(\dfrac{c+1}{a^2+1}\)>=3
\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{ab^2+b^2}{b^2+1}\) minh hong biet phai lam gi tiep theo dau
Bài 1
\(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}=a-\dfrac{a^2}{a+1}+b-\dfrac{b^2}{b+1}+c-\dfrac{c^2}{c+1}\)
\(=1-\left(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\right)\)
Áp dụng bđt Cauchy dạng phân thức \(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+3}=\dfrac{1}{1+3}=\dfrac{1}{4}\)
\(\Rightarrow1-\left(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}\right)\le1-\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow GTLN=\dfrac{3}{4}\) Dấu ''='' xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Bài 2
\(P=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}=\dfrac{a}{b^2+1}+\dfrac{1}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{1}{c^2+1}+\dfrac{c}{a^2+1}+\dfrac{1}{a^2+1}\)
Xét \(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{a^2c}{a^2+1}\)
Xét \(\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}+\dfrac{1}{a^2+1}=1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}+1-\dfrac{a^2}{a^2+1}\)
\(\Rightarrow P=6-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}+\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Áp dụng bđt Cauchy cho 2 số thực dương ta có \(b^2+1\ge2b\Rightarrow\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ac}{2}\)
Theo hệ quả của bđt Cauchy ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow3\ge ab+bc+ac\) \(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ac}{2}\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{3}{2}\)
Áp dụng bđt Cauchy cho 2 số thực dương ta có \(a^2+1\ge2a\Rightarrow\dfrac{a^2}{a^2+1}\le\dfrac{a^2}{2a}=\dfrac{a}{2}\)
\(\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Rightarrow P\ge6-\left(\dfrac{3}{2}+\dfrac{3}{2}\right)=3\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c=1\)
Bài 1 : Ta có : \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}=\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}+\dfrac{c^2}{c^2+c}\)
Theo BĐT CÔ - SI dưới dạng engel ta có :
\(\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}+\dfrac{c^2}{c^2+c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+\left(a+b+c\right)}=\dfrac{1}{a^2+b^2+c^2+1}\le\dfrac{1}{\dfrac{1}{a+b+c}+1}=\dfrac{1}{\dfrac{1}{3}+1}=\dfrac{4}{3}\)
Híc híc rối nùi luôn rồi , chắc sai ...
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Có BĐT: \(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Ta có:
\(VT=\)\(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\)
\(=\dfrac{1+1+c^2}{\left(a^2+b^2+1\right)\left(1+1+c^2\right)}+\dfrac{1+1+a^2}{\left(b^2+c^2+1\right)\left(1+1+a^2\right)}+\dfrac{1+1+b^2}{\left(c^2+a^2+1\right)\left(1+1+b^2\right)}\)
Áp dụng BĐT Bunhiacopski cho mẫu số, ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+c^2\right)\ge\left(a+b+c\right)^2\)
\(\left(b^2+c^2+1\right)\left(1+1+a^2\right)\ge\left(b+c+a\right)^2\)
\(\left(c^2+a^2+1\right)\left(1+1+b^2\right)\ge\left(c+a+b\right)^2\)
\(\Rightarrow VT\le\dfrac{1+1+c^2}{\left(a+b+c\right)^2}+\dfrac{1+1+a^2}{\left(b+c+a\right)^2}+\dfrac{1+1+b^2}{\left(c+a+b\right)^2}=\dfrac{6+a^2+b^2+c^2}{\left(a+b+c\right)^2}\le\dfrac{6+ab+bc+ca}{3\left(ab+bc+ca\right)}=\dfrac{6+3}{3.3}=1\)
\("="\Leftrightarrow a=b=c=1\)
Chắc đề bị nhầm rồi.
\(\dfrac{a}{\sqrt{b+1}}+\dfrac{b}{\sqrt{c+1}}+\dfrac{c}{\sqrt{a+1}}\ge2\sqrt{2}\left(\dfrac{a}{3+b}+\dfrac{b}{3+c}+\dfrac{c}{3+a}\right)\)
\(\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(ab+bc+ca\right)}\ge2\sqrt{2}.\dfrac{9}{9+\dfrac{\left(a+b+c\right)^2}{3}}=2\sqrt{2}.\dfrac{9}{12}=\dfrac{3}{\sqrt{2}}\)
Bài 1:
\(P=(x+1)\left(1+\frac{1}{y}\right)+(y+1)\left(1+\frac{1}{x}\right)\)
\(=2+x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}\)
Áp dụng BĐT Cô-si:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(x+\frac{1}{2x}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
\(y+\frac{1}{2y}\geq 2\sqrt{\frac{1}{2}}=\sqrt{2}\)
Áp dụng BĐT SVac-xơ kết hợp với Cô-si:
\(\frac{1}{2x}+\frac{1}{2y}\geq \frac{4}{2x+2y}=\frac{2}{x+y}\geq \frac{2}{\sqrt{2(x^2+y^2)}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Cộng các BĐT trên :
\(\Rightarrow P\geq 2+2+\sqrt{2}+\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
Vậy \(P_{\min}=4+3\sqrt{2}\Leftrightarrow a=b=\frac{1}{\sqrt{2}}\)
Bài 2:
Áp dụng BĐT Svac-xơ:
\(\frac{1}{a+3b}+\frac{1}{b+a+2c}\geq \frac{4}{2a+4b+2c}=\frac{2}{a+2b+c}\)
\(\frac{1}{b+3c}+\frac{1}{b+c+2a}\geq \frac{4}{2b+4c+2a}=\frac{2}{b+2c+a}\)
\(\frac{1}{c+3a}+\frac{1}{c+a+2b}\geq \frac{4}{2c+4a+2b}=\frac{2}{c+2a+b}\)
Cộng theo vế và rút gọn :
\(\Rightarrow \frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\geq \frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
a) Câu này biến đổi tương đương
b)
Ta có : \(a^2\left(a-1\right)^2\left(2+a\right)\ge0\Leftrightarrow a^2\left(3a-a^3-2\right)\le0\)
\(\Leftrightarrow3a^3+6-a^5-2a^2\le6\Leftrightarrow\left(3-a^2\right)\left(a^3+2\right)\le6\)
\(\Leftrightarrow\dfrac{1}{a^3+2}\ge\dfrac{3-a^2}{6}\)
Tương tự với b , c ta có :
\(\sum\left(\dfrac{1}{a^3+2}\right)\ge\sum\left(\dfrac{3-a^2}{6}\right)=\dfrac{9-\sum a^2}{6}=1\)
Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)
Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)
\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)
\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)
Áp dụng bđt Cauchy ta có :
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Thiết lập tương tự và thu lại ta có :
\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=3\)
Áp dụng BĐT cauchy ngược dấu ta có:
\(\dfrac{1}{a^2+1}=1-\dfrac{a^2}{a^2+1}\ge1-\dfrac{a^2}{2a}=1-\dfrac{a}{2}\)
Chứng minh tương tự ta có:
\(\dfrac{1}{b^2+1}\ge1-\dfrac{b}{2};\dfrac{1}{c^2+1}\ge1-\dfrac{c}{2}\)
Từ đó ta có: \(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge1-\dfrac{a}{2}+1-\dfrac{b}{2}+1-\dfrac{c}{2}=\)\(=3-\dfrac{a+b+c}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\left(đpcm\right)\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\) ≥ \(\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+3}=\dfrac{9}{a^2+b^2+c^2+3}\left(1\right)\)
Ta có BĐT : \(a^2+b^2+c^2\text{≥}ab+bc+ac\)
⇔ \(3\left(a^2+b^2+c^2\right)\text{≥}\left(a+b+c\right)^2\)
⇔ \(a^2+b^2+c^2\text{≥}\dfrac{9}{3}=3\left(2\right)\)
Từ ( 1 ; 2 ) ⇒ đpcm .
"=" ⇔ \(a=b=c=\dfrac{1}{3}\)