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\(A=\dfrac{ab+10b+25}{ab+5a+5b+25}+\dfrac{bc+10c+25}{bc+5b+5c+25}+\dfrac{ca+10a+25}{ac+5a+5c+25}\)
\(=\dfrac{\left(ab+5b\right)+\left(5b+25\right)}{\left(ab+5a\right)+\left(5b+25\right)}+\dfrac{\left(bc+5c\right)+\left(5c+25\right)}{\left(bc+5b\right)+\left(5c+25\right)}+\dfrac{\left(ca+5a\right)+\left(5a+25\right)}{\left(ac+5a\right)+\left(5c+25\right)}\)
\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{a\left(c+5\right)+5\left(c+5\right)}\)
\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)
\(=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)
\(=\left(\dfrac{b}{b+5}+\dfrac{5}{b+5}\right)+\left(\dfrac{a}{a+5}+\dfrac{5}{a+5}\right)+\left(\dfrac{c}{c+5}+\dfrac{5}{c+5}\right)\)
\(=1+1+1=3\) (\(a;b;c\ne-5\))
\(A=\dfrac{ab+5b+5b+25}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{bc+5c+5c+25}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{ca+5a+5a+25}{a\left(c+5\right)+5\left(c+5\right)}\)
\(A=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)
\(A=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)
\(A=\dfrac{a+5}{a+5}+\dfrac{b+5}{b+5}+\dfrac{c+5}{c+5}=1+1+1=3\)
Với dự đoán P đạt Min tại \(a=b=c=\frac{5}{3}\Rightarrow P=\frac{9}{20}\). Nên ta chứng minh \(P\ge\frac{9}{20}\).Thật vậy:\(P=\Sigma\frac{a}{ab+5c}=\Sigma\frac{a}{\left(a+c\right)\left(b+c\right)}=\frac{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\frac{\left(a+b+c\right)^2-\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{\left(a+b+c\right)^2-\frac{\left(a+b+c\right)^2}{3}}{\left[\frac{\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}{3}\right]^3}=\frac{9}{20}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{5}{3}\)
Vậy..
Lời giải:
Xét \(a^3+b^3-ab(a+b)=(a+b)(a-b)^2\geq 0, \forall a,b>0\)
Do đó \(a^3+b^3\geq ab(a+b)\) với mọi $a,b>0$
\(\Rightarrow b^3\geq ab(a+b)-a^3\)
\(\Rightarrow \frac{5a^3-b^3}{ab+3a^2}\leq \frac{5a^3-[ab(a+b)-a^3]}{ab+3a^2}=\frac{6a^2-b(a+b)}{b+3a}\)
hay \(\frac{5a^3-b^3}{ab+3a^2}\leq \frac{(2a-b)(3a+b)}{b+3a}=2a-b\)
Hoàn toàn tương tự ta có:
\(\frac{5b^3-c^3}{bc+3b^2}\leq 2b-c; \frac{5c^3-a^3}{ca+3c^2}\leq 2c-a\)
Cộng theo vế các BĐT thu được:
\(\text{VT}\leq a+b+c\leq 2018\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{2018}{3}\)
Lời giải:
Bạn nhớ tới bổ đề sau: Với $a,b>0$ thì $a^3+b^3\geq ab(a+b)$.
Áp dụng vào bài:
$5a^3-b^3\leq 5a^3-[ab(a+b)-a^3]=6a^3-ab(a+b)$
$\Rightarrow \frac{5a^3-b^3}{ab+3a^2}\leq \frac{6a^3-ab(a+b)}{ab+3a^2}=\frac{6a^2-ab-b^2}{3a+b}=\frac{(3a+b)(2a-b)}{3a+b}=2a-b$
Tương tự:
$\frac{5b^3-c^3}{bc+3b^2}\leq 2b-c; \frac{5c^3-a^3}{ca+3c^2}\leq 2c-a$
Cộng theo vế:
$\Rightarrow \text{VT}\leq a+b+c=3$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
Lời giải:
Do $a+b+c=5$ nên:
$Q=\frac{a}{ab+c(a+b+c)}+\frac{b}{bc+a(a+b+c)}+\frac{c}{ca+b(a+b+c)}=\frac{a}{(c+b)(c+a)}+\frac{b}{(a+b)(a+c)}+\frac{c}{(b+c)(b+a)}$
$=\frac{a(a+b)+b(b+c)+c(c+a)}{(a+b)(b+c)(c+a)}$
Theo BĐT AM-GM:
$(a+b)(b+c)(c+a)\leq \left(\frac{a+b+b+c+c+a}{3}\right)^3=\left(\frac{2(a+b+c)}{3}\right)^3=\frac{1000}{27}$
Và:
$a(a+b)+b(b+c)+c(c+a)=(a+b+c)^2-(ab+bc+ac)\geq (a+b+c)^2-\frac{(a+b+c)^2}{3}=\frac{50}{3}$
Do đó:
$Q\geq \frac{\frac{50}{3}}{\frac{1000}{27}}=\frac{9}{20}$
Vậy $Q_{\min}=\frac{9}{20}$. Dấu "=" xảy ra khi $a=b=c=\frac{5}{3}$
Lời giải:
Do $a+b+c=5$ nên:
$Q=\frac{a}{ab+c(a+b+c)}+\frac{b}{bc+a(a+b+c)}+\frac{c}{ca+b(a+b+c)}=\frac{a}{(c+b)(c+a)}+\frac{b}{(a+b)(a+c)}+\frac{c}{(b+c)(b+a)}$
$=\frac{a(a+b)+b(b+c)+c(c+a)}{(a+b)(b+c)(c+a)}$
Theo BĐT AM-GM:
$(a+b)(b+c)(c+a)\leq \left(\frac{a+b+b+c+c+a}{3}\right)^3=\left(\frac{2(a+b+c)}{3}\right)^3=\frac{1000}{27}$
Và:
$a(a+b)+b(b+c)+c(c+a)=(a+b+c)^2-(ab+bc+ac)\geq (a+b+c)^2-\frac{(a+b+c)^2}{3}=\frac{50}{3}$
Do đó:
$Q\geq \frac{\frac{50}{3}}{\frac{1000}{27}}=\frac{9}{20}$
Vậy $Q_{\min}=\frac{9}{20}$. Dấu "=" xảy ra khi $a=b=c=\frac{5}{3}$
\(R=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{1}=9\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
Vậy GTNN của \(R\) là \(9\) khi \(a=b=c=\frac{1}{3}\)
Chúc bạn học tốt ~
Ta chứng minh bổ đề sau:
\(\dfrac{5b^3-a^3}{ab+3b^2}\le2b-a\)
\(\Leftrightarrow5b^3-a^3\le\left(2b-a\right)\left(ab+3b^2\right)\)
\(\Leftrightarrow5b^3-a^3\le2ab^2+6b^3-a^2b-3b^2a\)
\(\Leftrightarrow a^3+b^3-a^2b-b^2a\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-2ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)
Bất đẳng thức cuối luôn đúng, vậy ta có
\(M\le2a-b+2b-c+2c-a=a+b+c\)Chứng minh hoàn tất. Đẳng thức xảy ra khi \(a=b=c\)
\(\dfrac{a}{ab+bc+ac+c^2}=\dfrac{a}{\left(a+c\right)\left(b+c\right)}\)
\(\dfrac{b}{bc+ac+ab+a^2}=\dfrac{b}{\left(a+b\right)\left(a+c\right)}\)
\(\dfrac{c}{ac+ab+b^2+bc}=\dfrac{c}{\left(a+b\right)\left(b+c\right)}\)
tiep di bn ơi