Cho a,b,c>0 và a+b+c=3
Tìm Bmin= \(\dfrac{a}{ab+a}+\dfrac{b}{bc+1}+\dfrac{c}{ca+1}\)
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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 9 2021
\(P=\dfrac{a^3}{b^2+ab+bc+ca}+\dfrac{b^3}{c^2+ab+bc+ca}+\dfrac{c^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}\)
Ta có:
\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge\dfrac{3a}{4}\)
\(\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3b}{4}\)
\(\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge\dfrac{3c}{4}\)
Cộng vế:
\(P+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\left(a+b+c\right)\ge\dfrac{1}{4}.\sqrt{3\left(ab+bc+ca\right)}=\dfrac{\sqrt{3}}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 4 2021
Do \(a;b;c\in\left[0;1\right]\Rightarrow\left(1-a\right)\left(1-c\right)\ge0\)
\(\Leftrightarrow ac+1\ge a+c\)
\(\Rightarrow1+b+ac\ge a+b+c\Rightarrow\dfrac{1}{1+b+ac}\le\dfrac{1}{a+b+c}\)
Tương tự: \(\dfrac{1}{1+c+ab}\le\dfrac{1}{a+b+c}\) ; \(\dfrac{1}{1+a+bc}\le\dfrac{1}{a+b+c}\)
Cộng vế với vế:
\(\dfrac{1}{1+b+ca}+\dfrac{1}{1+c+ab}+\dfrac{1}{1+a+bc}\le\dfrac{3}{a+b+c}\) (đpcm)
AH
Akai Haruma
Giáo viên
14 tháng 1 2023
Lời giải:
\(P=\frac{3}{ab+bc+ac}+\frac{5}{(a+b+c)^2-2(ab+bc+ac)}=\frac{3}{ab+bc+ac}+\frac{5}{1-2(ab+bc+ac)}\)
\(=\frac{3}{x}+\frac{5}{1-2x}\) với $x=ab+bc+ac$
Theo BĐT AM-GM:
$1=(a+b+c)^2\geq 3(ab+bc+ac)$
$\Rightarrow x=ab+bc+ac\leq \frac{1}{3}$
Vậy ta cần tìm min $P=\frac{3}{x}+\frac{5}{1-2x}$ với $0< x\leq \frac{1}{3}$
Áp dụng BĐT Bunhiacopxky:
$(\frac{3}{x}+\frac{5}{1-2x})[2x+(1-2x)]\geq (\sqrt{6}+\sqrt{5})^2$
$\Leftrightarrow P\geq (\sqrt{6}+\sqrt{5})^2=11+2\sqrt{30}$
Vậy $P_{\min}=11+2\sqrt{30}$
Giá trị này đạt tại $x=3-\sqrt{\frac{15}{2}}$
28 tháng 5 2021
Áp dụng bđt Cauchy - Schwarz ta có:\(Q=\dfrac{2-2a^2b^2}{\left(1+a^2\right)\left(1+b^2\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2\left(1-ab\right)\left(1+ab\right)}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2\left(bc+ca\right)\left(1+ab\right)}{\left(a+b\right)^2\left(b+c\right)\left(c+a\right)}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2c\left(1+ab\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2c\left(1+ab\right)}{\sqrt{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}+\dfrac{2}{\sqrt{1+c^2}}\le\dfrac{2c\left(1+ab\right)}{\sqrt{\left(ab+1\right)^2\left(c^2+1\right)}}+\dfrac{2}{\sqrt{1+c^2}}=\dfrac{2c}{\sqrt{c^2+1}}+\dfrac{2}{\sqrt{c^2+1}}=\dfrac{2\left(c+1\right)}{\sqrt{c^2+1}}\le\dfrac{2\left(c+1\right)}{\sqrt{\dfrac{\left(c+1\right)^2}{2}}}=2\sqrt{2}\)Dấu "=" xảy ra khi a = b = \(\sqrt{2}-1;c=1\).
Vậy..
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
\(\dfrac{a+b}{ab+c^2}=\dfrac{\left(a+b\right)^2}{\left(ab+c^2\right)\left(a+b\right)}=\dfrac{\left(a+b\right)^2}{b\left(a^2+c^2\right)+a\left(b^2+c^2\right)}\le\dfrac{a^2}{b\left(a^2+c^2\right)}+\dfrac{b^2}{a\left(b^2+c^2\right)}\)
Tương tự:
\(\dfrac{b+c}{bc+a^2}\le\dfrac{b^2}{c\left(a^2+b^2\right)}+\dfrac{c^2}{b\left(a^2+c^2\right)}\) ; \(\dfrac{c+a}{ca+b^2}\le\dfrac{c^2}{a\left(b^2+c^2\right)}+\dfrac{a^2}{c\left(a^2+b^2\right)}\)
Cộng vế:
\(VT\le\dfrac{1}{a}\left(\dfrac{b^2}{b^2+c^2}+\dfrac{c^2}{b^2+c^2}\right)+\dfrac{1}{b}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{c^2}{a^2+c^2}\right)+\dfrac{1}{c}\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
24 tháng 6 2021
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
AT
24 tháng 6 2021
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
HT
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 8 2021
Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)
\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)
đề thế này à bạn nãy mình đọc chắc nhầm
\(\Sigma\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge\Sigma a-\dfrac{a^2b}{2\sqrt{ab}}=\Sigma a-\dfrac{a\sqrt{ab}}{2}\)
\(\Rightarrow\Sigma\dfrac{a}{ab+1}\ge a+b+c-(\dfrac{a\sqrt{ab}}{2}+\dfrac{b\sqrt{bc}}{2}+\dfrac{c\sqrt{ca}}{2})\)
\(=3-\dfrac{1}{2}\left(\Sigma a\sqrt{ab}\right)\ge3-\dfrac{1}{2}\left(\dfrac{a\left(a+b\right)}{2}+\dfrac{b\left(b+c\right)}{2}+\dfrac{c\left(c+a\right)}{2}\right)=3-\dfrac{1}{4}\left(a^2+b^2+c^2+ab+bc+ca\right)=3-\dfrac{1}{4}\left[\left(a+b+c\right)^2-\left(ab+bc+ca\right)\right]\ge3-\dfrac{1}{4}\left[3^2-\dfrac{\left(a+b+c\right)^2}{3}\right]=3-\dfrac{1}{4}\left[3^2-\dfrac{3^2}{3}\right]=\dfrac{3}{2}\)
\(dấu"="\Leftrightarrow a=b=c=1\)
\(B=\Sigma\dfrac{1}{b+1}\ge\dfrac{9}{a+b+c+1+1+1}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
\(dấu"="\Leftrightarrow a=b=c=1\)