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\(BDT\Leftrightarrow\dfrac{\dfrac{1}{a}+\dfrac{1}{a^2}}{1+\dfrac{1}{a}+\dfrac{1}{a^2}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{b^2}}{1+\dfrac{1}{b}+\dfrac{1}{b^2}}+\dfrac{\dfrac{1}{c}+\dfrac{1}{c^2}}{1+\dfrac{1}{c}+\dfrac{1}{c^2}}\le2\)
Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(n,h,t\right)\) thì ta có :
\(\Leftrightarrow\dfrac{n+n^2}{1+n+n^2}+\dfrac{h+h^2}{1+h+h^2}+\dfrac{t+t^2}{1+t+t^2}\le2\)
\(\Leftrightarrow\dfrac{1}{1+n+n^2}+\dfrac{1}{1+h+h^2}+\dfrac{1}{1+t+t^2}\ge1\)
Đặt \(n=\dfrac{yz}{x^2};h=\dfrac{xz}{y^2};t=\dfrac{xy}{z^2}\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\)
Và \(\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+xy^2z+x^2z^2}+\dfrac{z^4}{z^4+xyz^2+x^2y^2}\ge1\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\)
Cần cm \(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2}\ge1\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)
\(\Leftrightarrow x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)\ge x^4+y^4+z^4+x^2yz+xy^2z+xyz^2+x^2y^2+y^2z^2+z^2x^2\)
\(\Leftrightarrow x^2y^2+y^2z^2+z^2x^2\ge x^2yz+xy^2z+xyz^2\left(1\right)\)
Áp dụng BĐT AM-GM ta có:
\(x^2y^2+y^2z^2=y^2\left(x^2+z^2\right)\ge2xy^2z\)
Tương tự rồi cộng theo vế ta có \(\left(1\right)\) đúng
Khi \(a=b=c=1\)
Sửa đề\(VP\le 2\) sau đó nó chính là 1 dạng của BĐT Vasc k cần thêm j cả :">
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:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\text{≥}\) \(\left(a+b\right)ab\)
⇒ \(a^3+b^3+abc\text{≥}\left(a+b\right)ab+abc=ab\left(a+b+c\right)\)
Tương tự : \(b^3+c^3+abc\text{ ≥}\left(b+c\right)bc+abc=bc\left(a+b+c\right)\)
\(c^3+a^3+abc\text{ ≥}\left(a+c\right)ac+abc=ac\left(a+b+c\right)\)
⇒ \(VT\text{ }\text{≤}\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\)
Đặt \(\left\{{}\begin{matrix}a^2-1=x\\b^2-1=y\\c^2-1=z\end{matrix}\right.\)(x,y,z>0)thì giả thiết trở thành \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
chứng minh \(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{y+1}+1}+\dfrac{1}{\sqrt{z+1}+1}\le1\)
Áp dụng BĐT cauchy:(dạng \(\dfrac{1}{a+b+c}\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\))(
\(\sum\dfrac{1}{\sqrt{x+1}+1}\le\sum\dfrac{1}{9}\left(\dfrac{4}{\sqrt{x+1}}+1\right)=\dfrac{4}{9}\left(\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}}\right)+\dfrac{1}{3}\)(*)
mà theo BĐT bunyakovsky:\(\left(\sum\dfrac{1}{\sqrt{x+1}}\right)^2\le3\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\le\dfrac{3}{16}\left(\dfrac{9}{x}+\dfrac{9}{y}+\dfrac{9}{z}+3\right)=\dfrac{3}{16}\left(9+3\right)=\dfrac{9}{4}\)
\(\Leftrightarrow\sum\dfrac{1}{\sqrt{x+1}}\le\dfrac{3}{2}\)kết hợp với (*), ta có
\(VT\le\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{3}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
Dấu = xảy ra khi x=y=z=3 hay a=b=c=2 (a,b,c>1)
Sửa đề: C/m: \(\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)
Ta có: \(a,b,c>1:\)
\(\Rightarrow\left\{{}\begin{matrix}a-1>0\\b-1>0\\c-1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a-1}>0\\\dfrac{1}{b-1}>0\\\dfrac{1}{c-1}>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}>0\)
Quay lại bài toán, theo giả thiết ta có:
\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}-\dfrac{1}{a-1}-\dfrac{1}{b-1}-\dfrac{1}{c-1}=1\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c-1}=1+\dfrac{1}{a-1}+\dfrac{1}{b-1}+\dfrac{1}{c-1}\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}-\dfrac{1}{c+1}< 1\)(đpcm)
Áp dụng bất đẳng thức Cauchy-Schwarz:\(\left\{{}\begin{matrix}\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}\ge\dfrac{\left(1+1\right)^2}{a+2b+c+c+3a}=\dfrac{4}{4a+2b+2c}=\dfrac{2}{c+2a+b}\\\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}\ge\dfrac{\left(1+1\right)^2}{b+2c+a+a+3b}=\dfrac{4}{4b+2c+2a}=\dfrac{2}{a+2b+c}\\\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{\left(1+1\right)^2}{c+2a+b+b+3c}=\dfrac{4}{4c+2a+2b}=\dfrac{2}{b+2c+a}\end{matrix}\right.\)
Cộng theo vế ta có:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}+\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}+\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{2}{c+2a+b}+\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}\)
Hay \(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\le\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\left(đpcm\right)\)
Áp dụng BĐT Cô si dạng Engel ; ta có :
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}\ge\dfrac{\left(1+1\right)^2}{\left(a+2b+c\right)+\left(c+3a\right)}=\dfrac{4}{4a+2b+2c}=\dfrac{2}{2a+b+c}\\ \dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}\ge\dfrac{\left(1+1\right)^2}{\left(b+2c+a\right)+\left(a+3b\right)}=\dfrac{4}{4b+2c+2a}=\dfrac{2}{2b+c+a}\\ \dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{\left(1+1\right)^2}{\left(c+2a+b\right)+\left(b+3c\right)}=\dfrac{4}{4c+2a+2b}=\dfrac{2}{2c+a+b}\)
\(\Rightarrow\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}+\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\\ \Rightarrow\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
Đề có bị sai không bạn theo mình thì phải là \(\ge8\) mới đúng
Áp dụng bất đẳng thức cô si cho hai số thực không âm ta có :
\(\dfrac{a^2}{b-1}+4\left(b-1\right)\ge2\sqrt{\dfrac{a^2}{b-1}\times4\left(b-1\right)}=4a\) (1)
\(\dfrac{b^2}{a-1}+4\left(a-1\right)\ge2\sqrt{\dfrac{b^2}{a-1}\times4\left(a-1\right)}=4b\) (2)
Cộng (1) và (2) vế theo vế ,ta được :
\(\dfrac{a^2}{b-1}+\dfrac{b^2}{a-1}+4a+4b-8\ge4a+4b\)
\(\Rightarrow\dfrac{a^2}{b-1}+\dfrac{b^2}{a-1}\ge8\)
Dấu "="xảy ra khi:a=b=2
Vậy \(\dfrac{a^2}{b-1}+\dfrac{b^2}{a-1}\ge8\) với a>1,b>1
Lời giải:
Do $abc=1$ nên tồn tại $x,y,z>0$ sao cho:
\((a,b,c)=\left(\frac{x^2}{yz}, \frac{y^2}{xz}, \frac{z^2}{xy}\right)\)
Khi đó: \(\text{VT}=\frac{1}{\frac{x^2}{yz}+\frac{y^2}{xz}+1}+\frac{1}{\frac{y^2}{xz}+\frac{z^2}{xy}+1}+\frac{1}{\frac{x^2}{yz}+\frac{z^2}{xy}+1}\)
\(\Leftrightarrow \text{VT}=\frac{xyz}{x^3+y^3+xyz}+\frac{xyz}{y^3+z^3+xyz}+\frac{xyz}{z^3+x^3+xyz}\)
Áp dụng BĐT Cô -si: \(\left\{\begin{matrix} x^3+y^3+y^3\geq 3xy^2\\ x^3+x^3+y^3\geq 3x^2y\end{matrix}\right.\)
\(\Rightarrow 3(x^3+y^3)\geq 3xy(x+y)\Leftrightarrow x^3+y^3\geq xy(x+y)\)
\(\Rightarrow x^3+y^3+xyz\geq xy(x+y+z)\)
\(\Rightarrow \frac{xyz}{x^3+y^3+xyz}\leq \frac{xyz}{xy(x+y+z)}=\frac{z}{x+y+z}\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(\text{VT}\leq \frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=\frac{x+y+z}{x+y+z}=1\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)