Cho a,b >0 thỏa mãn \(a+b\ge2.\)Tìm GTLN của \(M=\dfrac{1}{a+b^2}+\dfrac{1}{b+a^2}\)
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Y
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 3 2021
\(\dfrac{1}{1+a}=1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\)
Tương tự:
\(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\) ; \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+c\right)}}\)
Nhân vế với vế:
\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)
\(\Rightarrow abc\le\dfrac{1}{8}\)
\(N_{max}=\dfrac{1}{8}\) khi \(a=b=c=\dfrac{1}{2}\)
XO
11 tháng 2 2022
Ta có \(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}=2\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{\sqrt{ab}}=4\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=4-\dfrac{2}{\sqrt{ab}}\)
Khi đó P = \(\dfrac{1}{\sqrt{ab}}\left(4-\dfrac{2}{\sqrt{ab}}\right)=-2\left(\dfrac{1}{\sqrt{ab}}-1\right)^2+2\le2\)
Dấu "=" khi a = b = 1
HX
2
3 tháng 4 2022
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
XO
30 tháng 12 2021
\(4M=\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{4}{\left(c+a\right)+\left(b+c\right)}\)
\(\le\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{b+c}\)
\(=\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
=> 8M \(\le\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}=2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=8\)
=> \(M\le1\)
Dấu "=" xảy ra <=> a = b = c = 3/4
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2021
\(\dfrac{1}{2a+b+c}=\dfrac{1}{a+a+b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Tương tự:
\(\dfrac{1}{a+2b+c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)
Cộng vế:
\(M\le\dfrac{1}{16}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1\)
\(M_{max}=1\) khi \(a=b=c=\dfrac{3}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2023
\(Q=\dfrac{2a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{a+c}.\dfrac{c}{2\left(b+c\right)}}\)
\(\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{a+c}+\dfrac{c}{2\left(b+c\right)}\right)\)
\(=\dfrac{9}{4}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\)
SW
1
AH
Akai Haruma
Giáo viên
1 tháng 6 2021
Lời giải:
$1=a+b+3ab\leq (a+b)+3.\frac{(a+b)^2}{4}$
$\Rightarrow a+b\geq \frac{2}{3}$
$\Rightarrow a^2+b^2\geq \frac{(a+b)^2}{2}=\frac{2}{9}$
\(p=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1-(a+b)}{a+b}=\sqrt{1-a^2}+\sqrt{1-b^2}+\frac{1}{a+b}-1\)
\(\leq \sqrt{(1-a^2+1-b^2)(1+1)}+\frac{1}{\frac{2}{3}}-1=\sqrt{2(2-a^2-b^2)}+\frac{1}{2}\)
Mà \(2-a^2-b^2\leq 2-\frac{2}{9}=\frac{16}{9}\)
Do đó:
\(P\leq \sqrt{\frac{32}{9}}+\frac{1}{2}=\frac{3+8\sqrt{2}}{6}\) và đây chính là giá trị max.
AH
Akai Haruma
Giáo viên
1 tháng 6 2021
SKY WARS:
Đặt $a+b=t$ thì:
$1\leq t+\frac{3}{4}t^2$
$\Leftrightarrow 4\leq 4t+3t^2$
$\Leftrightarrow 3t^2+4t-4\geq 0$
$\Leftrightarrow (3t-2)(t+2)\geq 0$
Vì $t>0$ nên $3t-2\geq 0\Rightarrow t\geq \frac{2}{3}$
BC
11 tháng 3 2021
Nè Phan Linh Nhi, mk ko hỉu cái chỗ: a+b\(\le2\). Bn có thể giải thích chi tiết cho mk đc ko??
\(\left(a+b^2\right)\left(a+1\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{1}{a+b^2}\le\dfrac{a+1}{\left(a+b\right)^2}\)
Tương tự: \(\dfrac{1}{b+a^2}\le\dfrac{b+1}{\left(a+b\right)^2}\)
\(\Rightarrow M\le\dfrac{a+b+2}{\left(a+b\right)^2}=\dfrac{2}{\left(a+b\right)^2}+\dfrac{1}{a+b}=\dfrac{2}{\left(a+b\right)^2}+\dfrac{1}{a+b}-1+1\)
\(\Rightarrow M\le\left(\dfrac{2}{a+b}-1\right)\left(\dfrac{1}{a+b}+1\right)+1=\left(\dfrac{2-a-b}{a+b}\right)\left(\dfrac{1}{a+b}+1\right)+1\le1\)
\(M_{max}=1\) khi \(a=b=1\)