# Inequality Only with Am-Gm

I want to prove that if $$ab+bc+ca=3$$ for any $$a,b,c>0$$ real number, then

$$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \geq 3\sqrt{2}.$$

I read solution that used Cauchy-Schwarz inequality, but I want to prove this claim using only Am-Gm inequality.

I tried to prove this statement: If $$a>b \geq c$$, then $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} > 3\sqrt{2}.$$ With this claim we can prove that for $$a=b=c$$ we have a minimum, so $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}=3\sqrt{2}$$ We have $$3(a+b+c)^2 \geq 3(ab+bc+ca) \geq 9$$ so $$(a+b+c) \geq 3$$, and $$1 \geq abc$$.

If $$a>b$$:$$a=b+k$$, $$k>0$$
So we have $$\sqrt{b+k}+\sqrt{b+1}+\sqrt{c+1}$$. Do you think this kind of idea can lead to anything?

• I have two different solutions, but no one of them by AM-GM or C-S. I am ready to show, if you want. Aug 10 at 10:24
• Contradiction method might be useful Aug 10 at 10:43
• @MichaelRozenberg I dont have any doubt! I want try a little bit more. Thank you Aug 10 at 11:32
• @MichaelRozenberg write here your solutions! Aug 11 at 12:55
• @MichaelRozenberg thank you Aug 11 at 18:21

By Holder $$\left(\sum_{cyc}\sqrt{a+1}\right)^2\sum_{cyc}\frac{(a+2)^3}{a+1}\geq(a+b+c+6)^3$$ and it's enough to prove that: $$(a+b+c+6)^3\geq18\sum_{cyc}\frac{(a+2)^3}{a+1}.$$ Now, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$,where $$v>0$$, and $$abc=w^3$$.

Thus, after homogenization we need to prove that: $$(3u+6v)^3\geq18v\sum_{cyc}\frac{(a+2v)^3}{a+v},$$ which is equivalent to $$f(w^3)\geq0$$, where $$f$$ is a linear function.

But the linear function gets a minimal value for an extremal value of $$w^3$$,

which by $$uvw$$ happens in the following cases.

1. $$w^3\rightarrow0^+0$$.

Let $$c\rightarrow0^+0$$ and $$b\rightarrow\frac{3}{a},$$ where $$a>0$$.

Thus, we obtain:$$\left(a+\frac{3}{a}+6\right)^3\geq18\left(\frac{(a+2)^3}{a+1}+\frac{\left(\frac{3}{a}+2\right)^3}{\frac{3}{a}+1}+8\right)$$ or$$a^8+4a^7+30a^6+18a^5-162a^4+54a^3+270a^2+108a+81\geq0,$$ which is true by AM-GM: $$a^8+4a^7+30a^6+18a^5-162a^4+54a^3+270a^2+108a+81>$$ $$>\left(2\sqrt{30\cdot270}-162\right)a^4>0;$$ 2. Two variables are equal.

Let $$b=a$$ and $$c=\frac{3-a^2}{2a},$$ where $$0

Thus, we need to prove that: $$\left(2a+\frac{3-a^2}{2a}+6\right)^3\geq18\left(\frac{2(a+2)^3}{a+1}+\frac{\left(\frac{3-a^2}{2a}+2\right)^3}{\frac{3-a^2}{2a}+1}\right)$$ or $$(a-1)^2(3+8a+45a^2+24a^3-7a^4-a^6)\geq0$$ and we are done!

Proof.

Firstly, we'll prove the following problem as a lemma.

Problem. For all $$x,y,z\ge 0$$ satisfying $$x^2y^2+y^2z^2+z^2x^2+3(xyz)^2\ge 6,$$then we have $$x+y+z\ge 3.$$

(The proof is in the next part).

Now, denoting $$x=\sqrt{\frac{a+1}{2}};y=\sqrt{\frac{b+1}{2}};z=\sqrt{\frac{c+1}{2}},$$which the lemma says that it's enough to prove $$\frac{1}{4}\sum_{cyc}(a+1)(b+1)+\frac{3}{8}(a+1)(b+1)(c+1)\ge 6.$$ Let $$a+b+c=p\ge 3; 0 \le abc=r\le 1.$$ The inequality becomes $$2(6+2p)+3(p+r+4)\ge 48,$$or $$7p+3r\ge 24. \tag{*}$$ We split the inequality into two cases

• $$p\ge \dfrac{24}{7}.$$ The $$(*)$$ is obviously true.
• $$3\le p\le \dfrac{24}{7}.$$ By using Schur of third degree, $$r\ge \dfrac{p(12-p^2)}{9},$$ it remains to prove $$7p+\dfrac{p(12-p^2)}{3}-24\ge 0\iff \frac{(p-3)(-p^2-3p+24)}{3}\ge 0,$$which is true for all $$p\in\left[3;\dfrac{24}{7}\right].$$

Hence, the proof is done.

Also see here for the main idea of proof.

Can you go further by proving the lemma ?

• Very interesting your post about CY3. Thank you for sharing! Aug 11 at 22:23
• Did you complete it? Sep 11 at 2:24
• If you want i can give you an cauchy-schartz inequality Oct 7 at 6:49
• Would you please? Oct 8 at 0:49
• $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\geq \frac{3^{\frac{3}{2}}.\sqrt{a+b+c+3}}{3}=\sqrt{3(a+b+c+3)}\geq \sqrt{3(\sqrt{3(ab+bc+ca)}+3)}=\sqrt{3.6}=3\sqrt{2}$$ Oct 17 at 15:26

Alternative proof.

We can prove the following stronger inequality: for all $$a, b, c > 0$$, $$\left(\sqrt{a+1} + \sqrt{b+1} + \sqrt{c+1}\right)^2 \ge 18 + \frac{9(ab + bc + ca - 3)}{2(a + b + c)}. \tag{1}$$

Let $$x = \sqrt{a + 1} - 1, y = \sqrt{b + 1} - 1, z = \sqrt{c + 1} - 1$$. Then $$x, y, z \ge 0$$ and $$a = x^2 + 2x, b = y^2 + 2y, c = z^2 + 2z$$. (1) is written as \begin{align*} &\left(x + 1 + y + 1 + z + 1\right)^2 \\[6pt] \ge{}& 18 + \frac{9[(x^2 + 2x)(y^2 + 2y) + (y^2 + 2y)(z^2 + 2z) + (z^2 + 2z)(x^2 + 2x) - 3]}{2(x^2 + 2x + y^2 + 2y + z^2 + 2z)}. \tag{2} \end{align*}

We use the pqr method.

Let $$p = x + y + z, q = xy + yz + zx, r = xyz$$.

(2) is written as $$(p + 3)^2 \ge 18 + \frac{18pq + 9q^2 + 36q - 27 - (18p + 54)r}{2p^2 - 4q + 4p}. \tag{3}$$

From (3), using $$r \ge \frac{4pq - p^3}{9}$$ (degree three Schur), it suffices to prove that $$(p + 3)^2 \ge 18 + \frac{18pq + 9q^2 + 36q - 27 - (18p + 54)\cdot \frac{4pq - p^3}{9} }{2p^2 - 4q + 4p}$$ or (after clearing the denominators) $$f(q) := -9q^2 + (4p^2 - 18p)q + 10p^3 + 6p^2 - 36p + 27 \ge 0. \tag{4}$$

Note that $$0 \le q \le p^2/3$$ and $$f(q)$$ is concave. We have $$f(0) = 10p^3 + 6p^2 - 36p + 27 > 0$$ and $$f(p^2/3) = \frac13(p^2 + 6p - 9)^2 \ge 0$$. Thus, $$f(q) \ge 0$$ on $$[0, p^2/3]$$.

We are done.

• +1, nice! How did you come up with the idea proving stronger inequality? It's even non homogenous. Sep 11 at 2:27
• @TATAbox It is a way to eliminate the constraints. It is similar to this. Sep 11 at 2:32
• It seems interesting. Now I remember that Dragonheart6 did post this kind of method on AOPS. It is series of similar problem solved immediately by eliminate the constrains Sep 11 at 2:35
• @TATAbox Yes, it is helpful sometimes. Sep 11 at 2:36

My attempt: I want to prove that if $$ab+bc+ca=3$$ for any $$a,b,c>0$$ real number, then $$\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \geq 3\sqrt{2}$$ We put $$a+1=x^2, b+1=y^2, c+1=z^2$$, if we exploit that fact that $$ab+bc+ca=3$$ with the substitution we obtain: $$x^2y^2+y^2z^2+z^2x^2=2(x^2+y^2+z^2)$$ We focus on $$p(x)=x^2(y^2−2)+y^2(z^2−2)+z^2(x^2−2)$$ We have solution real solution if the discriminant is major or equal to 0:
$$δ=b^2−4ac=0^2−4ac=−4(y^2+z^2−2)(y^2)(z^2−2)(−z^2)=4(y^2+z^2−2)(y^2)(z^2)(z^2−2)$$ We have two cases: $$\begin{cases} (z^2−2)≥0\\ (y^2+z^2−2)≥0\\ \end{cases}$$ or $$\begin{cases} (z^2−2) \le 0\\ (y^2+z^2−2) \le 0\\ \end{cases}$$.
The second case is impossible because $$x^2=1,y^2=1$$ but we obtain $$a=0,b=0$$
The first case we obtain that $$z \geq \sqrt{2}$$ If we consider the $$p(y)$$ and $$p(z)$$ with the same argument we obtain $$y≥\sqrt{2}$$, and $$x\geq \sqrt{2}$$. So $$x+y+z≥3\sqrt{2}$$

• $x^2y^2+y^2z^2+z^2x^2=2(x^2+y^2+z^2)$ and $x, y, z \ge 1$ together implies $x + y + z \ge 3\sqrt 2$. Aug 12 at 2:07
• So it's true, i will try. Thank you Aug 12 at 13:08
• Look forward to your solution. Aug 12 at 13:12
• We note that $\sum_{cyc} x^2y^2 \geq 12$. We note that equality holds for $x=y=z$. So for $x=y=z=\sqrt{2}$ we have a minimum of the expression. It's not enough for to prove that $x+y+z \geq 3\sqrt2$?, because we have proved that we have a minimum for $x=y=z$. But it looks much easy in this way Aug 12 at 14:43
• @RiverLi I'am afriad the require some inequality that i dont study. I will return later to the problem Aug 14 at 23:14

Another way.

Let $$c\rightarrow0^+$$.

Thus, $$b\rightarrow\frac{3}{a}$$ and by Minkowski(triangle inequality) and AM-GM $$\sum_{cyc}\sqrt{a+1}=\sqrt{a+1}+\sqrt{\frac{3}{a}+1}+1\geq$$ $$\geq\sqrt{\left(\sqrt{a}+\sqrt{\frac{3}{a}}\right)^2+4}+1\geq\sqrt{4\sqrt3+4}+1>3\sqrt2.$$

Let $$f(a,b,c,\lambda)=\sum_{cyc}\sqrt{a+1}-3\sqrt2+\lambda(ab+ac+bc-3).$$ Thus, in the inside critical point $$(a,b,c)$$ should be: $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$$ or $$\frac{1}{2\sqrt{a+1}}+\lambda(b+c)=\frac{1}{2\sqrt{b+1}}+\lambda(a+c)=\frac{1}{2\sqrt{c+1}}+\lambda(a+b)=0.$$ Let in this point $$a\neq b$$ and $$a\neq c$$.

Thus, $$\frac{1}{2\sqrt{a+1}}+\lambda(b+c)=\frac{1}{2\sqrt{b+1}}+\lambda(a+c)$$ gives $$\frac{a-b}{\sqrt{(a+1)(b+1)}\left(\sqrt{a+1}+\sqrt{b+1}\right)}+2\lambda(a-b)=0$$ or $$\frac{1}{\sqrt{(a+1)(b+1)}\left(\sqrt{a+1}+\sqrt{b+1}\right)}+2\lambda=0.$$ By the similar way from $$\frac{1}{2\sqrt{a+1}}+\lambda(b+c)=\frac{1}{2\sqrt{c+1}}+\lambda(a+b)$$ we'll obtain: $$\frac{1}{\sqrt{(a+1)(c+1)}\left(\sqrt{a+1}+\sqrt{c+1}\right)}+2\lambda=0,$$ which gives $$\tfrac{1}{\sqrt{(a+1)(b+1)}\left(\sqrt{a+1}+\sqrt{b+1}\right)}=\tfrac{1}{\sqrt{(a+1)(c+1)}\left(\sqrt{a+1}+\sqrt{c+1}\right)}$$ or $$\sqrt{(b+1)(a+1)}+b+1=\sqrt{(a+1)(c+1)}+c+1$$ or $$(b-c)\sum_{cyc}\sqrt{a+1}=0,$$ which gives $$b=c$$ and it's enough to prove our inequality for equality case of two variables.

Let $$b=a$$ and $$c=\frac{3-a^2}{2a},$$ where $$0

Thus, we need to prove that:$$2\sqrt{a+1}+\sqrt{\frac{3-a^2}{2a}+1}\geq3\sqrt2$$ or after squaring of the both sides $$4(a+1)\sqrt{2a(3-a)}+7a^2-26a+3\geq0$$ or $$7(a-1)^2-4\left(3a+1-(a+1)\sqrt{2a(3-a)}\right)\geq0$$ or $$7(a-1)^2-\frac{4(a-1)^2(2a^2+2a+1)}{3a+1+(a+1)\sqrt{2a(3-a)}}\geq0,$$ for which it's enough to prove that: $$7(3a+1)-4(2a^2+2a+1)\geq0,$$ which is true because $$7(3a+1)-4(2a^2+2a+1)=13a+3-8a^2>$$ $$>13a-7a^2=a(13-7a)>a(13-7\sqrt3)>0$$ and we are done!

Another way.

Method of Lagrange's multipliers give the system: $$cyc: \frac1{2\sqrt{a+1}}=\lambda (b+c)$$ and by substracting cyclicly the system: $$cyc: (b-a)=2\lambda(b-a)\sqrt{a+1}\sqrt{b+1}(\sqrt{a+1}+\sqrt{b+1})$$ This system, together with the constraint $$ab+bc+ca=3$$ has the only inner solution $$a=b=c=1$$.

So, it is enough to check the boundary. Without loss of generality let $$c\to0$$. Then $$a=b=\sqrt3$$ is optimal and $$2\sqrt{\sqrt3 +1}+1\geq 3\sqrt2$$ is true. We are done.