CHO abcd=1. Tính:
A=\(\dfrac{a}{abc+ab+a+1}+\dfrac{b}{bcd+bc+b+1}+\dfrac{c}{cda+cd+c+1}+\dfrac{d}{dab+da+d+1}\)
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Bài 1: Ta có:
\(M=\frac{ad}{abcd+abd+ad+d}+\frac{bad}{bcd.ad+bc.ad+bad+ad}+\frac{c.abd}{cda.abd+cd.abd+cabd+abd}+\frac{d}{dab+da+d+1}\)
\(=\frac{ad}{1+abd+ad+d}+\frac{bad}{d+1+bad+ad}+\frac{1}{ad+d+1+abd}+\frac{d}{dab+da+d+1}\)
$=\frac{ad+abd+1+d}{ad+abd+1+d}=1$
Bài 2:
Vì $a,b,c,d\in [0;1]$ nên
\(N\leq \frac{a}{abcd+1}+\frac{b}{abcd+1}+\frac{c}{abcd+1}+\frac{d}{abcd+1}=\frac{a+b+c+d}{abcd+1}\)
Ta cũng có:
$(a-1)(b-1)\geq 0\Rightarrow a+b\leq ab+1$
Tương tự:
$c+d\leq cd+1$
$(ab-1)(cd-1)\geq 0\Rightarrow ab+cd\leq abcd+1$
Cộng 3 BĐT trên lại và thu gọn thì $a+b+c+d\leq abcd+3$
$\Rightarrow N\leq \frac{abcd+3}{abcd+1}=\frac{3(abcd+1)-2abcd}{abcd+1}$
$=3-\frac{2abcd}{abcd+1}\leq 3$
Vậy $N_{\max}=3$
cái này giống Câu hỏi của ngonhuminh - Toán lớp 10 | Học trực tuyến
ta có : \(M=\dfrac{1}{abc+ab+a+1}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\)
\(\Leftrightarrow M=\dfrac{abcd}{abcd+abc+ab+a}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{bcd}{bcd+bc+b+1}+\dfrac{1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{bcd+1}{bcd+bc+b+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abcd+bcd}{abcd+bcd+bc+b}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{acd+cd}{acd+cd+c+1}+\dfrac{1}{acb+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{acd+cd+1}{acd+cd+c+1}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abcd+acd+cd}{abcd+acd+cd+c}+\dfrac{1}{abd+ad+d+1}\) \(\Leftrightarrow M=\dfrac{abd+ad+d}{abd+ad+d+1}+\dfrac{1}{abd+ad+d+1}\)
Áp dụng BĐT Svacxơ:
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\ge\dfrac{4}{ab+bc+cd+da}\)
Áp dụng BĐT Cô-si:
\(\dfrac{4}{ab+bc+cd+da}\ge\dfrac{4}{a^2+b^2+c^2+d^2}\)
Ta cần c/m: \(\dfrac{4}{a^2+b^2+c^2+d^2}\ge a^2+b^2+c^2+d^2\)
\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)^2\ge4\)
Áp dụng BĐT Svacxơ: \(\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{c^2}{1}+\dfrac{d^2}{1}\right)^2\ge\dfrac{\left(a+b+c+d\right)^{2^2}}{16}\)
mà a+b+c+d=4 nên: \(\dfrac{\left(a+b+c+d\right)^4}{16}\ge\dfrac{64}{16}=4=VP\)
Vậy ta có đpcm.
\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ba}{abcd+abc+ab+a}+\dfrac{\dfrac{c}{cd}}{\dfrac{acd}{cd}+\dfrac{cd}{cd}+\dfrac{c}{cd}+\dfrac{1}{cd}}+\dfrac{\dfrac{d}{d}}{\dfrac{dab}{d}+\dfrac{ad}{d}+\dfrac{d}{d}+\dfrac{1}{d}}\)
\(A=\dfrac{a}{abc+ab+a+1}+\dfrac{ab}{1+abc+ab+a}+\dfrac{\dfrac{1}{d}}{a+1+\dfrac{1}{d}+\dfrac{1}{cd}}+\dfrac{1}{ab+a+1+\dfrac{1}{d}}\)
Mà \(abcd=1\Rightarrow\dfrac{1}{d}=abc;\dfrac{1}{cd}=ab\)
\(\Rightarrow A=\dfrac{a}{abc+ab+a+a}+\dfrac{ab}{abc+ab+a+1}+\dfrac{abc}{a+1+abc+ab}+\dfrac{1}{ab+a+1+abc}\)
\(\Rightarrow A=\dfrac{a+ab+abc+1}{abc+ab+a+1}=1\)