$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\geq \frac{2}{3}$ (a,b,c,d>0)
Lời giải. Áp dụng Cauchy-Schwarz ta có $$\sum \frac{a}{b+2c+3d}= \sum \frac{a^2}{ab+2ac+3ad} \ge \frac{(a+b+c+d)^2}{4(ab+bc+ca+ad+bd+cd)}$$
Bây giờ ta cần chứng minh $4(ab+bc+ca+ad+bd+cd) \le \frac{3}{2}(a+b+c+d)^2 \Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2+(a-d)^2+(c-d)^2+(b-d)^2 \ge 0$.
Dấu đẳng thức xảy ra khi và chỉ khi $a=b=c=d$. $\square$
$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 1$ (a,b,c>0)
Lời giải. Cũng áp dụng Cauchy-Shwarz ta có $$\sum \frac{a}{b+2c}= \sum \frac{a^2}{ab+2ac} \ge \frac{(a+b+c)^2}{3(ab+bc+ca)} \ge 1$$
Dấu đẳng thức xảy ra khi và chỉ khi $a=b=c$.
$\frac{9}{a+b+c}\leq \sum \frac{4}{2a+b+c}\leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ (a,b,c>0)
Lời giải. Áp dụng BĐT Cauchy-Schwarz ta có $$\sum \frac{4}{2a+b+c} \ge \frac{36}{4(a+b+c)}= \frac{9}{a+b+c}$$
Ta chứng minh vế còn lại: Áp dụng Cauchy-Schwarz ta có $$\begin{array}{l} \frac{4}{2a}+ \frac{1}{b}+ \frac{1}{c} \ge \frac{16}{2a+b+c} \\ \frac{4}{2b}+ \frac 1c + \frac 1a \ge \frac{16}{2b+c+a} \\ \frac{4}{2c}+ \frac 1a + \frac 1b \ge \frac{16}{2c+a+b} \end{array}$$
Do đó $4 \left( \frac 1a+ \frac 1b + \frac 1c \right) \ge \sum \frac{16}{2a+b+c}$.
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
