1

Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that$$\frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}\le 1+\sqrt{2}.$$Equality holds at $(0,1,1).$

I've tried to use Jichen lemma .

Firstly, we rewrite the OP as$$1+\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}\ge \frac{1}{\sqrt{a+1}}+\frac{1}{\sqrt{b+1}}+\frac{1}{\sqrt{c+1}}.$$ By using the lemma, we will prove three following inequalities$$2\ge \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}.$$ $$\frac{5}{4}\ge \sum_{cyc}\frac{1}{a+1}\frac{1}{b+1}$$ $$\color{red}{\frac{1}{4}\ge \frac{1}{a+1}\frac{1}{b+1}\frac{1}{c+1}.}$$ The last inequality is wrong already which says that Jichen lemma is not appropriate.

I hope to see some good ideas. Thank you.

IraeVid
  • 3,216
TATA box
  • 1
  • 1
  • 5
  • 29

4 Answers4

2

If $c=0$, so $ab=1$ and by C-S $$\sum_{cyc}\frac{1}{\sqrt{1+a}}=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+1\leq\sqrt{2\left(\frac{1}{1+a}+\frac{1}{1+b}\right)}+1=$$ $$=\sqrt{\frac{2(2+a+b)}{2+a+b}}+1=\sqrt2+1.$$ Now, let $f(a,b,c,\lambda)=\sum\limits_{cyc}\frac{1}{\sqrt{1+a}}+\lambda(ab+ac+bc-1)$ and let $(a,b,c)$ be an inside maximal point.

Thus, in this point should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$$ or $$-\frac{1}{2\sqrt{(1+a)^3}}+\lambda(b+c)=-\frac{1}{2\sqrt{(1+b)^3}}+\lambda(a+c)=-\frac{1}{2\sqrt{(1+c)^3}}+\lambda(a+b)=0,$$ which gives $$(a+1)^3(b+c)^2=(b+1)^3(a+c)^2=(c+1)^3(a+b)^2.$$ Now, let in this point $a\neq b$ and $a\neq c$.

Thus, $$(a+1)^3(b+c)^2=(b+1)^3(a+c)^2$$ gives $$(a-b)(1-a-b+c+3c^2-3ab+3abc-c^2ab)=0$$ and $$(a+1)^3(b+c)^2=(c+1)^3(a+b)^2$$ gives $$(a-c)(1-a-c+b+3b^2-3ac+3abc-b^2ac)=0,$$ which gives $$1-a-c+b+3b^2-3ac+3abc-b^2ac=1-a-b+c+3c^2-3ab+3abc-c^2ab$$ or $$(b-c)(2+3a+3b+3c-abc)=0$$ and since by AM-GM $$abc\leq\sqrt{\left(\frac{ab+ac+bc}{3}\right)^3}=\frac{1}{3\sqrt3}<1,$$ we obtain $b=c$ and it's enough to prove our inequality for equality case of two variables.

Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a\leq1.$

Thus, it's enough to prove that: $$\frac{2}{\sqrt{1+a}}+\frac{1}{\sqrt{1+\frac{1-a^2}{2a}}}\leq1+\sqrt2$$ and the rest is smooth.

Can you end it now?

Michael Rozenberg
  • 194,933
  • (+1) Thank you. There is a little typo: $0<a\le 1.$ I can work with a variable function. In the proof, I see a new symbol to me. If it is available for you, I hope you can clear it help me. – TATA box Aug 01 '23 at 10:24
  • 1
    @TATA box If you say about $\frac{\partial f}{\partial a}$, so it's a derivative of $f$ by $a$. – Michael Rozenberg Aug 01 '23 at 10:39
  • thanks. I will research about it. It might be a theorem about derivative. – TATA box Aug 01 '23 at 10:43
  • 1
    Why In the inside maximal point the derivative is equal to $0$? – Michael Rozenberg Aug 01 '23 at 10:45
  • Yes, I feel confusing about that claim. – TATA box Aug 01 '23 at 10:46
  • 1
    A full reasoning is the following. A function, which continuous on compact, gets on this compact a maximum value. Compact in $\mathbb R^n$ it's a closed and bounded set, it's a set like segment $[a,b]$, but in our case it's like $[a,b]^3$. Read about compact. If $f$ is a continuous function on $[a,b]$, so there is $c\in[a,b],$ for which for any $x\in[a,b]$ we have $f(x)\leq f(c)$. Id est, $f$ gets on $[a,b]$ the maximal value. Also, if it happens for $a<c<b$ and $f$ is a differentiable function, so $f'(c)=0.$ The similar thing happens in $\mathbb R^3$. – Michael Rozenberg Aug 01 '23 at 11:05
  • I am appreciate your explaination. My current knowledge is not good enough but I will try my best. – TATA box Aug 01 '23 at 11:11
1

Proof.

We have, for all $x \ge 0$ $$\frac{1}{\sqrt{x + 1}} \le \frac{2\sqrt 2 - 1}{7} + \frac{24 + 22\sqrt 2}{49x + 35 + 28\sqrt 2}. \tag{1}$$ (Note: We have $\mathrm{RHS}^2 - \mathrm{LHS}^2 = \frac{(9 - 4\sqrt 2)x(x-1)^2}{(7x + 5 + 4\sqrt 2)^2(x + 1)}\ge 0$.)

Using (1), it suffices to prove that $$3 \cdot \frac{2\sqrt 2 - 1}{7} + \sum_{\mathrm{cyc}} \frac{24 + 22\sqrt 2}{49a + 35 + 28\sqrt 2} \le 1 + \sqrt 2$$ or $$\sum_{\mathrm{cyc}} \frac{2}{7a + 5 + 4\sqrt 2} \le \sqrt 2 - 1$$ or (after clearing the denominators) $$(21 + 14\sqrt 2)(ab + bc + ca) + (5 + 4\sqrt 2)(a + b + c) + 49abc - 31 - 22\sqrt 2 \ge 0. \tag{2}$$

Using $(a + b + c)^2 \ge 3(ab + bc + ca)$, we have $a + b + c \ge \sqrt 3$.

If $a + b + c \ge 2$, we have $$\mathrm{LHS}_{(2)} \ge 21 + 14\sqrt 2 + (5 + 4\sqrt 2)\cdot 2 - 31 - 22\sqrt 2 = 0.$$

If $\sqrt 3 \le a + b + c < 2$, using $abc \ge \frac{4(a + b + c)(ab + bc + ca) - (a + b + c)^3}{9}$ (degree three Schur inequality), we have \begin{align*} \mathrm{LHS}_{(2)} &\ge -\frac{49}{9}(a + b + c)^3 + (241/9 + \sqrt 2)(a + b + c) - 10 - 8\sqrt 2\\[6pt] &\ge 0. \end{align*}

We are done.

River Li
  • 37,323
  • (+1) Thank you for your interest. The estimate (1) is specific. I think you worked with function to find it. – TATA box Aug 01 '23 at 14:29
  • @TATAbox Simply find a bound of the form $\frac{1}{\sqrt{x + 1}} \le \frac{qx + r}{x + p}$ for some appropriate constants $p, q, r$. – River Li Aug 01 '23 at 14:31
  • I agree with you. Btw, https://math.stackexchange.com/questions/4743736/if-abbcca-1-prove-sqrt5a5b8ab-sqrt5c5b8cb-sqrt5a5c8ac-ge-3. Have you tried it? I hope we can find a good approach. – TATA box Aug 01 '23 at 14:36
  • @TATAbox I don't have a nice proof now. – River Li Aug 01 '23 at 14:37
  • @ River Li, Even it is ugly, could you please give me an advice ? – TATA box Aug 01 '23 at 14:39
  • 1
    @TATAbox You may try Lagrange Multiplier or KKT conditions. But I don't like them for Olympiad type inequalities. So I don't do that. – River Li Aug 01 '23 at 14:47
  • 1
    @River Li Nice solution! +1. We can use LM method in math Olympiad. Just we need to explain our steps. – Michael Rozenberg Aug 01 '23 at 17:42
  • You used some software/calculator to handle some of the inequalities that make up the bulk of your answer. I would bet you could never do this and "discover" the some inequality manually in <1 hour in competition. In this sense, there is no sentence as meaningless as "We are done" Because, in fact, your answer without using a calculator/software is just not useful, however it is correct or not . – User Aug 01 '23 at 21:39
  • @MichaelRozenberg Thanks. You can use LM. But I don't use LM in general. I agree with Yufei Zhao's words "In fact, calculus is best avoided in olympiad solutions as it is generally viewed unfavorably." – River Li Aug 01 '23 at 22:27
  • 1
    @User I am not interested in solutions which can be done during contests. I am interested in solutions which can be verified easily (e.g. by hand) but can be motivated by software. "We are done." just to tell that the proof ends. I am interested in difficult inequalities including open problems. Many inequalities do not have nice proofs. For example, in my answer, I used computer to find the SOS proof. However, it can be verified even by hand in less than one hour. – River Li Aug 01 '23 at 22:40
  • "It can be verified", but we need to find the solution , not verification . Some difficult inequalities posted here under the contest math tag Therefore, In general computer based proof of inequality is not a technique . – User Aug 01 '23 at 22:55
  • 1
    @MichaelRozenberg By the way, I think that TATAbox actually does not understand well your answer using derivatives, from his words "By the way, My current knowledge is not good enough but I will try my best." Sometimes, a proof without calculus is easier to understood for some people. – River Li Aug 01 '23 at 22:55
  • @User Sometimes, it is not indeed a contest problem. For the OP, I don't see a typical contest solution using classical or known inequalities such as C-S, AM-GM, etc. Even it is indeed a contest problem, I just provide a solution which can be verified by hand. Actually, in my many answers, such as this one, we can do it by hand (although I do it by computer). For example, I can find the bound $\frac{1}{\sqrt{x + 1}} \le \frac{2\sqrt 2 - 1}{7}
    • \frac{24 + 22\sqrt 2}{49x + 35 + 28\sqrt 2}$ by hand.
     – River Li Aug 01 '23 at 23:00
  • 1
    @User Here is one more example in which the solution is easily verified by hand but motivated by computer. Problem 1: Let $a, b, c$ be real numbers with $a + b \ge 0, b + c \ge 0, c + a\ge 0$. Prove that $a^3 + b^3 + c^3 - a^2b - b^2c - c^2a \ge 0$.

    My computer assisted proof: We have $$a^3 + b^3 + c^3 - a^2b - b^2c - c^2a = \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2

    • \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}.$$

    Although we used a computer to find it, it is easily verified by hand.

     – River Li Aug 01 '23 at 23:18
  • 2
    @River Li. I respect your contributions in forum. This problem is from my school's math team test. I posted it to find some nice ideas. I understand the approach given by Michael Rozenberg. The core idea is proving the OP is true in case two of equal variables. I tend to think the motivation of a proof. The theorem behind strange symbol is barrier to my current 10 th- grade knowledge. That why I wrote the comment. If I just apply Michale'idea without well-understanding, I will not do that. – TATA box Aug 02 '23 at 02:51
  • 2
    ....Some problems is hard with equality ocurring at a=b=c, we can use this theorem and contradiction method to manipulate a proof working with one variable function.About the words "Sometimes, a proof without calculus is easier to understood for some people." I agree with you in some cases. – TATA box Aug 02 '23 at 02:51
  • 1
    PROBLEM 2. For all non-negative real numbers $a,b,c$ such that $ab+bc+ca=1.$ Prove that $$\frac{1-a}{a+\sqrt{bc}}+\frac{1-b}{b+\sqrt{ca}}+\frac{1-c}{c+\sqrt{ab}}\ge 1.$$ This hard problem is my training team' test. There are some reasoning approachs. In my during time, I use $uvw$ after expanding all yields although it is ugly. The calculating is complicated without calculator but it is only idea I came up with. Obviously, I can not give full proof. – TATA box Aug 02 '23 at 03:15
  • 2
    ... The official solution is short- 1 line. Using Cauchy- Schwarz by multiplying $a+b+c.$ It is really confusing to me. Some short proofs is nice sight but the inside core is very hard to see. We discovered many proving inequality methods and the number of them will not stop. No matter we know how much technique, the key is using a suitable one. – TATA box Aug 02 '23 at 03:15
  • 1
    @TATAbox You may post it as another question. Perhaps some people are interested in it. It is not good to give here as a comment. – River Li Aug 02 '23 at 04:07
  • It was old. I gave it here as an example. – TATA box Aug 02 '23 at 04:09
  • Did you use Pade approximation for the $(1)$? – youthdoo Aug 02 '23 at 04:29
  • @youthdoo It is not Pade. But it is inspired or similar. We need to fit the equality case $x=0, 1$ by assuming the form $\frac{1}{\sqrt{x+1}} \le \frac{qx + r}{x + p}$. – River Li Aug 02 '23 at 06:02
0

Proof

By Cauchy-Schwarz \begin{align*} &\sum_{cyc}\frac{1}{\sqrt{1+x}}\le \sqrt{\sum_{cyc}{\frac{x+1+\sqrt{2}}{x+1}}\sum_{cyc}{\frac{1}{x+1+\sqrt{2}}}}\le 1+\sqrt{2}\\&\Leftrightarrow \sum_{\mathrm{cyc}}{\left(\begin{array}{c} 4x^3y^2z+\left( \dfrac{2\sqrt{2}}{3}+3 \right) x^2y^2z^2+\\ (x+y-z)^2\left( \left( \sqrt{2}+2 \right) x^2z^2+\left( \sqrt{2}+1 \right) y^2z^2 \right) +\left( 4\sqrt{2}+6 \right) xy^4z \end{array} \right)}\\&+4\left( 5\sqrt{2}+6 \right) xyz(xy+xz+yz)^{3/2}\ge 0 \end{align*} We end proof here.

Dragon boy
  • 1
  • 16
0

Source? I remember it belongs Tran Nk Trang. Here is just a sketch of author's idea.

  • Firstly, we prove the following lemma.

If $x,y,z\ge 0$ such that $x^2+y^2+z^2\le 2$ and $$x^2y^2+y^2z^2+z^2x^2+(2+2\sqrt{2})x^2y^2z^2\le \dfrac{7+2\sqrt{2}}{4}$$then $x+y+z\le 1+\sqrt{2}.$

  • Apply the lemma by setting $x=\dfrac{1}{\sqrt{a+1}};y=\dfrac{1}{\sqrt{b+1}};z=\dfrac{1}{\sqrt{c+1}}.$

We will prove two inequalities $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le 2. \tag{1}$$ $$a+b+c+5+2\sqrt{2}\le \frac{7+2\sqrt{2}}{4}(a+1)(b+1)(c+1). \tag{2}$$ Can you end it now?

user26857
  • 52,094
Dragon boy
  • 1
  • 16