cho a,b,c là các só thực dương thỏa mãn a +2b +3c =13
tìm GTNN của P = \(\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\)
K
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14 tháng 5 2021
Ta có: \(a+2b+3c=13\)
\(\Leftrightarrow\left(a-1\right)+2\left(b-1\right)+3\left(c-1\right)=7\)
Mà \(7^2=\left[\left(a-1\right)+2\left(b-1\right)+3\left(c-1\right)\right]^2\)
\(\le\left(1^2+2^2+3^2\right)\left[\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\right]\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge\frac{7}{2}\)
Dấu "=" xảy ra khi: \(a-1=\frac{b-1}{2}=\frac{c-1}{3}\Rightarrow\hept{\begin{cases}a=\frac{3}{2}\\b=2\\c=\frac{5}{2}\end{cases}}\)
30 tháng 4 2020
\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)
\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)
Áp dụng BĐT Cosi ta có:
\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)
Từ (1)(2)(3) ta có:
\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)
Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)
Dấu "=" xảy ra <=> a=b=c=1
12 tháng 2 2018
p \(\ge\)\(\frac{4}{a^2+b^2+2\left(a+b\right)}\) +\(\sqrt{\left(1+ab\right)^2}\) (bunhia và cosi)
=\(\frac{4}{a^2+b^2+2ab}+1+ab=\frac{4}{\left(a+b\right)^2}+a+b+1\)
do \(a+b=ab\le\frac{\left(a+b\right)^2}{4}\Rightarrow a+b\ge4\)
dạt a+b = t thì t>=4
cần tìm min \(\frac{4}{t^2}+t+1=\frac{4}{t^2}+\frac{t}{16}+\frac{t}{16}+\frac{7t}{8}+1\)
\(\ge3.\sqrt[3]{\frac{4}{t^2}.\frac{t}{16}.\frac{t}{16}}+\frac{7.4}{8}+1=\frac{21}{4}\)
dau = xay ra khi a=b=2
LV
13 tháng 8 2017
\(P=\frac{a^3}{\left(a+1\right)\left(b+1\right)}+\frac{b^3}{\left(b+1\right)\left(c+1\right)}+\frac{c^3}{\left(c+1\right)\left(a+1\right)}-1\)
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 8 2021
\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
24 tháng 8 2021
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
\(P=a^2-2a+b^2-2b+c^2-2c+3\)
\(P=\left(a^2+\dfrac{9}{4}\right)+\left(b^2+4\right)+\left(c^2+\dfrac{25}{4}\right)-2a-2b-2c-\dfrac{19}{2}\)
\(P\ge3a+4b+5c-2a-2b-2c-\dfrac{19}{2}\)
\(P\ge a+2b+3c-\dfrac{19}{2}=13-\dfrac{19}{2}=\dfrac{7}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};2;\dfrac{5}{2}\right)\)
Anh ;-; em chưa kịp làm :|