Cho $a,b,c$ là các số thực dương. Chứng minh rằng:
$\frac{a^2+bc}{a^2+(b+c)^2}+\frac{b^2+ca}{b^2+(c+a)^2}+\frac{c^2+ab}{c^2+(a+b)^2}\leq \frac{18(a^2+b^2+c^2)}{5(a+b+c)^2}$
Cho $a,b,c$ là các số thực dương. Chứng minh rằng:
$\frac{a^2+bc}{a^2+(b+c)^2}+\frac{b^2+ca}{b^2+(c+a)^2}+\frac{c^2+ab}{c^2+(a+b)^2}\leq \frac{18(a^2+b^2+c^2)}{5(a+b+c)^2}$
Cho $a,b,c$ là các số thực dương. Chứng minh rằng:
$\frac{a^2+bc}{a^2+(b+c)^2}+\frac{b^2+ca}{b^2+(c+a)^2}+\frac{c^2+ab}{c^2+(a+b)^2}\leq \frac{18(a^2+b^2+c^2)}{5(a+b+c)^2}$
BĐT đã cho tương đương với:
$1-\frac{a^2+bc}{a^2+(b+c)^2}+1-\frac{b^2+ca}{b^2+(c+a)^2}+1-\frac{c^2+ab}{c^2+(a+b)^2}\geq 3-\frac{18(a^2+b^2+c^2)}{5(a+b+c)^2}$
$\Leftrightarrow \sum \frac{b^2+c^2+bc}{a^2+(b+c)^2}\geq \frac{30(ab+bc+ca)-3(a^2+b^2+c^2)}{5(a+b+c)^2}$ (1)
Áp dụng BĐT AM-GM và BĐT Cauchy-Schwarz, ta có
$VT(1)\geq \sum \frac{3(b+c)^2}{4\left [ a^2+(b+c)^2 \right ]}\geq \frac{12(a+b+c)^2}{4(3(a^2+b^2+c^2)+2(ab+bc+ca))}=\frac{3(a+b+c)^2}{3(a^2+b^2+c^2)+2(ab+bc+ca)}$
Do đó để chứng minh (1) ta chỉ cần chứng minh
$\frac{(a+b+c)^2}{3(a^2+b^2+c^2)+2(ab+bc+ca)}\geq \frac{10(ab+bc+ca)-(a^2+b^2+c^2)}{5(a+b+c)^2}\Leftrightarrow 5(a+b+c)^4\geq \left [ 3(a^2+b^2+c^2)+2(ab+bc+ca) \right ]\left [ 10(ab+bc+ca)-(a^2+b^2+c^2) \right ]$
BĐT này đúng vì theo BĐT AM-GM, ta có $45VP=\left [ 50(ab+bc+ca)-5(a^2+b^2+c^2) \right ]\left [ 27(a^2+b^2+c^2)+18(ab+bc+ca) \right ]\leq \left [ \frac{22(a^2+b^2+c^2)+68(ab+bc+ca)}{2} \right ]^2=\left [ \frac{22(a+b+c)^2+24(ab+bc+ca)}{2} \right ]^2\leq \left [ \frac{22(a+b+c)^2+8(a+b+c)^2}{2} \right ]^2=225(a+b+c)^4=45VT$
(đpcm)
Bài viết đã được chỉnh sửa nội dung bởi vutuanhien: 10-06-2013 - 18:43
"The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!" - Grothendieck
Do đó để chứng minh (1) ta chỉ cần chứng minh
$\frac{(a+b+c)^2}{3(a^2+b^2+c^2)+2(ab+bc+ca)}\geq \frac{10(ab+bc+ca)-(a^2+b^2+c^2)}{5(a+b+c)^2}\Leftrightarrow 5(a+b+c)^4\geq \left [ 3(a^2+b^2+c^2)+2(ab+bc+ca) \right ]\left [ 10(ab+bc+ca)-(a^2+b^2+c^2) \right ]$
Bất đẳng thức này có thể dễ dàng chứng minh hơn nếu ta đặt $a^2+b^2+c^2=x,ab+bc+ca=y$ thì $x \ge y$.
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”).
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