Inequality Only with Am-Gm

Question

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 $\sqrt[2]{b+k}+\sqrt[2]{b+1}+\sqrt[2]{c+1}$. Do you think this kind of idea can lead to anything?

Answer 1

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 $0<a<\frac{3-a^2}{2a}.$

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!

Answer 2

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.

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Can you go further by proving the lemma ?

Answer 3

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.

Answer 4

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 $$x+y+z≥3\sqrt{2}$$

Answer 5

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 $0<a<\sqrt3.$

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!

Answer 6

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.