Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có \(-\dfrac{4ab^2}{4b^2+1}\ge-\dfrac{4ab^2}{2\sqrt{4b^2}}=\dfrac{4ab^2}{4b}=ab\)
\(-\dfrac{4a^2b}{4a^2+1}\ge-\dfrac{4a^2b}{2\sqrt{4a^2}}=\dfrac{4a^2b}{4a}=ab\)
Mà \(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4ab^2}{4a^2+1}\ge a-ab+b-ab=4ab-2ab=2ab\)
Mà \(a+b=4ab\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)
\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)
Dấu "=" \(\Leftrightarrow a=b=\dfrac{1}{2}\)
Lời giải:
ĐK $\Rightarrow \frac{1}{a}+\frac{1}{b}=4$
Đặt $\frac{1}{x}=a; \frac{1}{y}=b$ thì bài toán trở thành:
Cho $a,b>0$ thỏa mãn $a+b=4$. CMR:
$P=\frac{x^2}{y(x^2+4)}+\frac{y^2}{x(y^2+4)}\geq \frac{1}{2}$
-----------------------
Áp dụng BĐT AM-GM:
$\frac{x^2}{y(x^2+4)}+\frac{y(x^2+4)}{64}\geq \frac{x}{4}$
$\frac{y^2}{x(y^2+4)}+\frac{x(y^2+4)}{64}\geq \frac{y}{4}$
Cộng theo vế và rút gọn:
$P\geq \frac{3(x+y)-xy}{16}=\frac{12-xy}{16}$
Mà $xy\leq \frac{(x+y)^2}{4}=4$
$\Rightarrow P\geq \frac{12-4}{16}=\frac{1}{2}$
Ta có đpcm.
Đề bài sai
Đề đúng: \(\dfrac{1}{\sqrt{a}+2\sqrt{b}+3}+\dfrac{1}{\sqrt{b}+2\sqrt{c}+3}+\dfrac{1}{\sqrt{c}+2\sqrt{a}+3}\le\dfrac{1}{2}\)
\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)
\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)
Ta có: \(\frac{a}{1+4b^2}=\frac{a\left(1+4b^2\right)-4ab^2}{1+4b^2}=a-\frac{4ab^2}{1+4b^2}\ge a-\frac{4ab^2}{2\sqrt{4b^2.1}}=a-\frac{2ab^2}{2b}=a-ab\)(bđt cosi)
CMTT: \(\frac{b}{1+4a^2}\ge b-ab\)
=> P \(\ge a+b-2ab=4ab-2ab=2ab\)
Mặt khác ta có: \(a+b\ge2\sqrt{ab}\)(cosi)
=> \(4ab\ge2\sqrt{ab}\) <=> \(2ab\ge\sqrt{ab}\)<=> \(4a^2b^2-ab\ge0\) <=> \(ab\left(4ab-1\right)\ge0\)
<=> \(\orbr{\begin{cases}ab\le0\left(loại\right)\\ab\ge\frac{1}{4}\end{cases}}\)(vì a,b là số thực dương)
=> P \(\ge2\cdot\frac{1}{4}=\frac{1}{2}\)
Dấu "=" xảy ra <=> a = b = 1/2
Vậy MinP = 1/2 <=> a = b= 1/2
Ta có: \(a+b=4ab\le\left(a+b\right)^2\Leftrightarrow\left(a+b\right)\left[\left(a+b\right)-1\right]\ge0\)
Mà \(a+b>0\Rightarrow a+b\ge1\)
Áp dụng BĐT Cô-si, ta có: \(P=\frac{a}{1+4b^2}+\frac{b}{1+4a^2}=\left(a-\frac{4ab^2}{1+4b^2}\right)+\left(b-\frac{4a^2b}{1+4a^2}\right)\)\(\ge\left(a-\frac{4ab^2}{4b}\right)+\left(b-\frac{4a^2b}{4a}\right)=\left(a+b\right)-2ab=\left(a+b\right)-\frac{a+b}{2}=\frac{a+b}{2}\ge\frac{1}{2}\)
Đẳng thức xảy ra khi a = b = 1/2
\(VT=3\left(\dfrac{1}{4ab}+\dfrac{1}{a^2+4b^2}\right)+\dfrac{1}{2.a.2b}\ge\dfrac{12}{a^2+4ab+4b^2}+\dfrac{2}{\left(a+2b\right)^2}=14\)
Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)
* Vì \(a,b\ge1\)nên \(\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab+1\ge a+b\)
Một cách tương tự: \(bc+1\ge b+c;ca+1\ge c+a\)
Với mọi số thực \(a\ge1\) ta luôn có: \(\left(a-1\right)^2\ge0\Leftrightarrow a^2\ge2a-1\Leftrightarrow\frac{1}{2a-1}\ge\frac{1}{a^2}\)
Tương tự: \(\frac{1}{2b-1}\ge\frac{1}{b^2};\frac{1}{2c-1}\ge\frac{1}{c^2}\)
Từ đó suy ra \(VT\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{4ab}{ab+1}+\frac{4bc}{bc+1}+\frac{4ca}{ca+1}\)\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+4-\frac{4}{ab+1}+4-\frac{4}{bc+1}+4-\frac{4}{ca+1}\)\(\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}-\frac{4}{ab+1}-\frac{4}{bc+1}-\frac{4}{ca+1}+12\)\(\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}-\frac{4}{a+b}-\frac{4}{b+c}-\frac{4}{c+a}+12\)\(=\left(\frac{2}{a+b}-1\right)^2+\left(\frac{2}{b+c}-1\right)^2+\left(\frac{2}{c+a}-1\right)^2+9\ge9\)
Đẳng thức xảy ra khi a = b = c = 1
\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}=\dfrac{a\left(4b^2+1\right)}{4b^2+1}-\dfrac{4ab^2}{4b^2+1}+\dfrac{b\left(4a^2+1\right)}{4a^2+1}-\dfrac{4a^2b}{4b^2+1}\)
\(\ge a-\dfrac{4ab^2}{4b}+b-\dfrac{4a^2b}{4a}\) (bđt Cô-si)
=a-ab+b-ab=a+b-2ab=4ab-2ab=2ab
Lại có a+b=4ab \(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=4\ge\dfrac{2}{2\sqrt{ab}}\Rightarrow4\sqrt{ab}\ge2\Rightarrow ab\ge\dfrac{1}{4}\)
\(\Rightarrow2ab\ge\dfrac{1}{2}\Rightarrow\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)
Dấu ''='' xảy ra khi \(a=b=\dfrac{1}{2}\)
\(\dfrac{a}{4b^2+1}+\dfrac{b}{4a^2+1}\ge\dfrac{1}{2}\)
\(\Leftrightarrow a-\dfrac{a}{4b^2+1}+b-\dfrac{b}{4a^2+1}\le a+b-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{4ab^2}{4b^2+1}+\dfrac{4ba^2}{4a^2+1}\le4ab-\dfrac{1}{2}\)
\(\sum\dfrac{4ab^2}{4b^2+1}\le^{CS}2ab\)
\(\Rightarrow CM:2ab\le4ab-\dfrac{1}{2}\Leftrightarrow ab\ge\dfrac{1}{4}\)
Từ GT \(\Rightarrow4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)
\(\Rightarrow dpcm\)