5

Here is question 1 from USAMO 2018 Q1 (held in April):

Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$. Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$

This question is on symmetric polynomials. I recall as many facts as I can think of regarding inequalities. (This test was closed book).

  • AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.
  • quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.
  • The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.
  • I am still looking for other inequalities that might work.

It's tempting to race through this problem with the first solution that comes to mind. I'm especially interested in some kind of organizing principle.

cactus314
  • 24,438

3 Answers3

8

Since our inequality and condition are symmetric and homogeneous,

we can assume that $abc=1$ and $a\geq b\geq c$.

Thus, $a+b+c=4$ and we need to prove that $$2(ab+ac+bc)\geq a^2+b^2-3c^2$$ or $$(a+b+c)^2\geq2(a^2+b^2-c^2)$$ or $$8\geq a^2+b^2-c^2$$ or $$8\geq a^2+b^2-(4-a-b)^2$$ or $$8\geq a^2+b^2-(16+a^2+b^2-8a-8b+2ab)$$ or $$12-4(a+b)+ab\geq0$$ or $$12-4(4-c)+\frac{1}{c}\geq0$$ or $$(2c-1)^2\geq0.$$ Done!

Michael Rozenberg
  • 194,933
  • And the case $$a\le b\le c$$? – Dr. Sonnhard Graubner May 16 '18 at 12:45
  • In your case the solution is the same. Finally, we'll get $(2a-1)^2\geq0.$ This inequality is symmetric and we can assume that $a\geq b\geq c$. – Michael Rozenberg May 16 '18 at 12:46
  • Are these iff statements? Are you sure you can go from $(2c-1)^2 \geq 0$ to the inequality $2(ab+ac+bc) \geq a^2 + b^2 - 3c^2$ ? – cactus314 May 16 '18 at 12:47
  • Yes, of course. See my solution. I am ready to explain it more. – Michael Rozenberg May 16 '18 at 12:53
  • 2
    Very interesting equality case (under the assumption $a \geq b \geq c$): $$\left(a,b,c\right)=\left(\frac{7+\sqrt{17}}{4},\frac{7-\sqrt{17}}{4},\frac{1}{2}\right)\lambda,,$$ where $\lambda \geq 0$. – Batominovski May 16 '18 at 12:59
  • You can't really let abc=1, and then claim you've proved the inequality, because abc might not be 1, and in some cases that might cause problems (say, for instance, we had the square root of abc instead of the cube root). But if you let a=Ak, b=Bk, and c=Ck, where ABC=1, then you can trivially prove that the factor k does indeed cancel everywhere. – dovalojd May 16 '18 at 12:59
  • @cactus314 I added something. See now. – Michael Rozenberg May 16 '18 at 13:02
  • @dovalojd That's what Michael left for the reader to think about. This is a standard trick if an inequality is homogeneous. Michael did give a complete (albeit in a reverse order) proof. It is not fun to read a math solution with everything handed to you. – Batominovski May 16 '18 at 13:02
  • @Batominovski I just want to know whether everything is handed to me or not... 'application to other abc is an exercise for the reader' would also have made it a good answer, but it was presented as a proof and needed the edit that added 'it's homogeneous' to actually be one. – dovalojd May 16 '18 at 13:32
  • @dovalojd We have different viewpoints. I prefer solutions (that are indeed correct) with various gaps (of reasonable difficulties) that I have to fill in myself. It's more entertaining and good for the brain. I do not expect complete answers from anybody except from my students. I just want to point out that your definition of "complete proofs" or "good answers" on this site is vastly different from mine. – Batominovski May 16 '18 at 15:04
1

Without loss of generality, let $a\le b=ax\le c=axy, x\ge 1, y\ge 1$.

Then: $$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a+ax+axy=4\sqrt[3]{a(ax)(axy)} \Rightarrow 1+x+xy=4x^{\frac23}y^{\frac13} \qquad (1)$$ Also: $$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a^2+b^2+c^2=16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \qquad (2)$$ Plugging $(2)$ and then $(1)$ into the given inequality: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2 \overbrace{\Rightarrow}^{(2)}\\ 2(ab+bc+ca)+4a^2\ge 16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \Rightarrow \\ (a(ax)+(ax)(axy)+(axy)a)+a^2\ge 4\sqrt[3]{(a(ax)(axy))^2} \Rightarrow\\ x+x^2y+xy+1\ge 4x\sqrt[3]{xy^2} \overbrace{\Rightarrow}^{(1)}\\ 4x^{\frac23}y^{\frac13}+x^2y\ge 4x^{\frac43}y^{\frac23} \Rightarrow\\ \left(x^{\frac23}y^{\frac13}\right)^2-4\left(x^{\frac23}y^{\frac13}\right)+4\ge 0 \Rightarrow \\ \left(x^{\frac23}y^{\frac13}-2\right)^2\ge 0.$$

farruhota
  • 31,482
1

$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$=$ 4(ab+bc+ca)+ 4\min(a^2,b^2,c^2)-2(ab+bc+ca)$

Now, suppose $a≤b≤c$ without losing generality. Hence,

$ 2(ab+bc+ca)$+$ 4\min( a^2,b^2,c^2)$

=$ 4(ab+bc+ca)+ 4a^2-2(ab+bc+ca)$

= $4((a+b+c)a+bc)-2(ab+bc+ca)$

=$4(4a(abc)^{1/3})+bc)-2(ab+bc+ca)≥ 16(abc)^{2/3}-2(ab+bc+ca) =(a+b+c)^2-2(ab+bc+ca)=a^2+b^2+c^2$

Alapan Das
  • 936