Cho $a,b,c\geqslant 0,a+b+c=1$. Chứng minh:
a. $a^3 + b^3 +c^3 + 6abc \geqslant \frac{1}{4}$
b.$ 7(ab+bc+ca)\leqslant 2 + 9abc$
Bài viết đã được chỉnh sửa nội dung bởi phanquockhanh: 30-08-2013 - 22:14
cho $a,b,c\geqslant 0,a+b+c=1$. Chứng minh:
b.$ 7(ab+bc+ca)\leqslant 2 + 9abc$
Lời giải. Đặt $a+b+c=p, ab+bc+ca=q,abc=r$ thì $p=1$. Áp dụng BĐT Schur ta có $4pq \le p^3+9r \Leftrightarrow 7q \le \frac{7+63r}{4}$.
Lại có $\frac{7+63r}{4} \le 2+9r \Leftrightarrow 27r \le 1$, đúng vì $27abc \le (a+b+c)^3=1$.
Dấu đẳng thức xảy ra khi và chỉ khi $a=b=c= \frac 13$.
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”).
Bài viết đã được chỉnh sửa nội dung bởi Phạm Hữu Bảo Chung: 30-08-2013 - 23:41
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