cho a,b,c là các số dương thỏa mãn: a+b+c=1
cmr :\(\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)
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AH
Akai Haruma
Giáo viên
8 tháng 6 2021
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(a^2+b^2+1)(1+1+c^2)\geq (a+b+c)^2$
$\Rightarrow \frac{1}{a^2+b^2+1}\leq \frac{c^2+2}{(a+b+c)^2}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
$\text{VT}\leq \frac{a^2+b^2+c^2+6}{(a+b+c)^2}=\frac{a^2+b^2+c^2+6}{a^2+b^2+c^2+2(ab+bc+ac)}\leq \frac{a^2+b^2+c^2+6}{a^2+b^2+c^2+2.3}=1$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=1$
1N
cho ba số dương \(0\le a\le b\le c\le1\) CMR \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le2\)
1
25 tháng 5 2022
Vì \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)
Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)
Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)
Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)
Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)
AH
Akai Haruma
Giáo viên
3 tháng 2
Lời giải:
$\text{VT}=\frac{a(a+b+c)+bc}{b+c}+\frac{b(a+b+c)+ac}{a+c}+\frac{c(a+b+c)+ab}{a+b}$
$=\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}+\frac{(c+a)(c+b)}{a+b}$
Áp dụng BĐT AM-GM:
$\frac{(a+b)(a+c)}{b+c}+\frac{(b+a)(b+c)}{a+c}\geq 2\sqrt{(a+b)^2}=2(a+b)$
$\frac{(b+c)(b+a)}{a+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(b+c)^2}=2(b+c)$
$\frac{(a+b)(a+c)}{b+c}+\frac{(c+a)(c+b)}{a+b}\geq 2\sqrt{(c+a)^2}=2(a+c)$
Cộng các BĐT trên theo vế và thu gọn:
$\text{VT}\geq 2(a+b+c)=2$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 2 2021
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
26 tháng 10 2021
Sửa \(\le\) thành \(\ge\) nha bạn
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)
Ta có \(\dfrac{a^2}{a+bc}=\dfrac{a^3}{a^2+abc}=\dfrac{a^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{b^2}{b+ca}=\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}\\\dfrac{c^2}{c+ba}=\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\end{matrix}\right.\)
Áp dụng BĐT cosi:
\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3}{4}a\)
\(\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{b^3}{64}}=\dfrac{3}{4}b\)
\(\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}+\dfrac{b+c}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{c^3}{64}}=\dfrac{3}{4}c\)
Cộng VTV:
\(\Leftrightarrow VT+\dfrac{a+b}{8}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\\ \Leftrightarrow VT\ge\dfrac{3\left(a+b+c\right)}{4}-\dfrac{2\left(a+b+c\right)}{8}\\ \Leftrightarrow VT\ge\dfrac{a+b+c}{4}\)
Dấu \("="\Leftrightarrow a=b=c=3\)
AH
Akai Haruma
Giáo viên
13 tháng 5 2023
Thỏa mãn $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ hay $a+b+c=1$ vậy bạn?
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 4 2021
\(\Leftrightarrow\left(1+ab+bc+ca\right)\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(a+b\right)\left(b+c\right)}+\dfrac{1}{\left(a+c\right)\left(b+c\right)}\right)\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{ab+bc+ca}{abc}\)
Áp dụng BĐT quen thuộc:
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(ab+bc+ca\right)\left(a+b+c\right)=\dfrac{8}{9}\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{2\left(1+ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\)
Ta chỉ cần chứng minh:
\(\dfrac{9\left(1+ab+bc+ca\right)}{4\left(ab+bc+ca\right)}\le\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow4\left(ab+bc+ca\right)^2\ge9abc+9abc\left(ab+bc+ca\right)\)
Do \(3\left(ab+bc+ca\right)^2\ge9abc\left(a+b+c\right)=9abc\)
Nên ta chỉ cần chứng minh:
\(\left(ab+bc+ca\right)^2\ge9abc\left(ab+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\ge9abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
Hiển nhiên đúng do \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 1 2022
\(a^2+b^2-ab\ge\dfrac{1}{2}\left(a+b\right)^2-\dfrac{1}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\dfrac{1}{\sqrt{a^2-ab+b^2}}\le\dfrac{1}{\sqrt{\dfrac{1}{4}\left(a+b\right)^2}}=\dfrac{2}{a+b}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tương tự:
\(\dfrac{1}{\sqrt{b^2-bc+c^2}}\le\dfrac{1}{2}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{c^2-ca+a^2}}\le\dfrac{1}{2}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(P\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
AH
Akai Haruma
Giáo viên
28 tháng 10 2021
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\sum \sqrt{\frac{ab}{c+ab}}=\sum \sqrt{\frac{ab}{c(a+b+c)+ab}}=\sum \sqrt{\frac{ab}{(c+a)(c+b)}}\)
\(\leq \sum \frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)=\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)
Vậy $P_{\max}=\frac{3}{2}$ khi $a=b=c=\frac{1}{3}$
Ta có \(\dfrac{ab+c}{c+1}=\dfrac{ab+c\left(a+b+c\right)}{\left(a+c\right)+\left(b+c\right)}=\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+c\right)+\left(b+c\right)}\)
\(\Rightarrow VT=\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+c\right)+\left(b+c\right)}+\dfrac{\left(a+b\right)\left(b+c\right)}{\left(a+b\right)+\left(b+c\right)}+\dfrac{\left(a+c\right)\left(a+b\right)}{\left(a+b\right)+\left(a+c\right)}\)
Đặt \(\left\{{}\begin{matrix}a+c=x\\b+c=y\\a+b=z\end{matrix}\right.\) \(\Rightarrow x+y+z=2\)
\(\Rightarrow VT\Leftrightarrow\dfrac{xy}{x+y}+\dfrac{yz}{z+y}+\dfrac{xz}{x+z}\)
Áp dụng bất đẳng thức \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\dfrac{xy}{x+y}\le\dfrac{xy}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{y}{4}+\dfrac{x}{4}\)
Thiết lập tương tự và thu lại ta có
\(\Rightarrow VT\le\dfrac{2\left(x+y+z\right)}{4}=1\) ( đpcm )
\(\Leftrightarrow\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)