limes123 wrote: Problem 2 Prove that if the real numbers a, b and c satisfy $a^2 + b^2 + c^2 = 3$ then $\sum \frac {a^2}{2 + b + c^2}\geq \frac {(a + b + c)^2}{12}$. When does the inequality hold? Uhh.. I solved it the day after the competition.. A bit too late :/ Cauchy-Swartz gives: ${ (\sum \frac {a^2}{2 + b + c^2})(\sum 2 + b + c^2}) \ge (a + b + c)^2$ So we just have to prove: $\sum 2 + b + c^2 \le 12$ or $a + b + c \le 3$ which is obvious when $a^2 + b^2 + c^2 = 3$. ($a + b + c \le \sqrt {(1+1+1)(a^2 + b^2 + c^2)} = 3$)

My solution is the same but I thought that if a,b,c are real the proof would be more complicated... Any other solutions?

Aaaaa right no, my first idea was to use the Engel's form of CS and I didn't try to prove it in other way.

limes123 wrote: Aaaaa right no, my first idea was to use the Engel's form of CS and I didn't try to prove it in other way. A little off topic, but I am a bit curious Did you attend to the Baltic Way for Poland?

No. I found the problems on Polish math forum.

See here: http://www.balticway08.math.univ.gda.pl/files/tasks/BW2008english.pdf

CS Engel $$\frac{a^2}{2+b+c^2}+\frac{b^2}{2+c+a^2}+\frac{c^2}{2+a+b^2} \ge \frac{(a+b+c)^2}{6+a+b+c+a^2+b^2+c^2}.$$ So it suffices to prove that $$6+a+b+c+a^2+b^2+c^2 \le 12.$$ Note that $a^2+b^2+c^2=3$, then we only need to prove that $$a+b+c \le 3$$ . But $$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \le a^2+b^2+c^2+2(a^2+b^2+c^2)=3(a^2+b^2+c^2)=9.$$ Hence $a+b+c \le 3$ which completes the proof. I'm sorry for some mistakes... Can we use CS Engel for real numbers?

Raja Oktovin wrote: CS Engel $$\frac {a^2}{2 + b + c^2} + \frac {b^2}{2 + c + a^2} + \frac {c^2}{2 + a + b^2} \ge \frac {(a + b + c)^2}{6 + a + b + c + a^2 + b^2 + c^2}.$$ So it suffices to prove that $$6 + a + b + c + a^2 + b^2 + c^2 \le 12.$$ Note that $a^2 + b^2 + c^2 = 3$, then we only need to prove that $$a + b + c \le 3$$ . But $$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \le a^2 + b^2 + c^2 + 2(a^2 + b^2 + c^2) = 3(a^2 + b^2 + c^2) = 9.$$ Hence $a + b + c \le 3$ which completes the proof. I'm sorry for some mistakes... Can we use CS Engel for real numbers? I don't know what CS Engel is... But it looks very much like Cauchy-Swartz.. And of course you can use negative values! (But remember that the square of a negative number is a positive number, so the biggest side will _always_ be non-negative) It seems like your proof is the exact same as mine.. So take a look at mine post, limes123 question and my answer.

Which is actually stronger? Cauchy Engel or cauchy schwartz as such? I feel i have solved a lot of problems including this one using cs.

Aravind Srinivas L wrote: Which is actually stronger? Cauchy Engel or cauchy schwartz as such? I feel i have solved a lot of problems including this one using cs. What is Cauchy Engel ?

the one that raja oktavin used.

Aravind Srinivas L wrote: the one that raja oktavin used. How does that differ from normal cauchy schwartz? I've always know cauchy schwartz as: $(a_1^2+...+a_n^2)(b_1^2+...+b_n^2) \ge (a_1b_1 + ... +a_nb_n)^2$. Is Cauchy Schwartz = Cauchy Engel ???

Hey Mathias , this is Cauchy Engel $\sum_{i = 1}^n \frac {x_i}{y_i}\geqslant \frac { (\sum_{i = 1}^n x_i)^2}{\sum_{i = 1}^n x_iy_i}$ Very useful !

Pablo09 wrote: Hey Mathias , this is Cauchy Engel $\sum_{i = 1}^n \frac {x_i}{y_i}\geqslant \frac { (\sum_{i = 1}^n x_i)^2}{\sum_{i = 1}^n x_iy_i}$ Very useful ! Ok. So that's just cauchy swartz $a_i^2 = \frac{x_i}{y_i}$ and $b_i^2 = x_iy_i$. $\iff$ Cauchy Swartz = Cauchy Engel. QED

Very similar to the above, but

Using C-S It suffices to prove that $(\sum 2+b+c^2)\le 12$ or $a+b+c\le 3$ from C-S $3(a^2+b^2+c^2)\ge (a+b+c)^2$

This inequality is easy to prove we just have to apply cauchy Schwartz once then we need to prove a+b+c<= 3 now $a^2+b^2+c^2$>= $(a+b+c) ^2$\3 so result follows

Note that $\sum_{cyc} \frac{a^2}{2+b+c^2} \ge \frac{(a+b+c)^2}{9+a+b+c}$ by Titu's Lemma, so it suffices to show $12 \ge 9+a+b+c \implies 3 \ge a+b+c$. But by Cauchy we know $(1+1+1)(a^2+b^2+c^2)=9 \ge (a+b+c)^2$, so $3 \ge a+b+c$.

Can we use C-S since $a,b,c$ are real, not necessarily positive?

jasperE3 wrote: Can we use C-S since $a,b,c$ are real, not necessarily positive? I guessed everybody used that... Btw, is Engel's form of Cauchy Schwarz is Titu's Lemma?

jasperE3 wrote: Can we use C-S since $a,b,c$ are real, not necessarily positive? You could have checked this here instead of asking. SatisfiedMagma wrote: I guessed everybody used that... Btw, is Engel's form of Cauchy Schwarz is Titu's Lemma? And to answer your question yes.

But I mean C-S in Engel form.

Yes, since $a\ge -\sqrt{3}$, etc. hence the quantities $2+a+b^2$, etc. are all positive. It was also stated in #4.

By Titu's Lemma, $$\sum_{\text{cyc}}\frac{a^2}{2+b+c^2}\ge\frac{(a+b+c)^2}{6+a^2+b^2+c^2+a+b+c}\ge\frac{(a+b+c)}{12}$$ as Cauchy yields $(a+b+c)^2\le (1+1+1)(a^2+b^2+c^2)=9.$ $\square$