$\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \geq \sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)} \geq 3\sqrt[3]{(a+1)(b+1)(c+1)}=6$

I understand how we get $\sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)} \geq 3\sqrt[3]{(a+1)(b+1)(c+1)}$ from the AM-GM inequality, and how we get $3\sqrt[3]{(a+1)(b+1)(c+1)}=6$ based on what's given to us in the problem, but I don't really get how the solution above gets $a^2+b^2+2\geq(a+1)(b+1).$ It seems obvious, but I can't prove it.

You know that $\frac{a^2+b^2}{2} \geq ab$, right? So you just need to show that $\frac{a^2+b^2}{2} + 1 \geq a + b$. Try using some squares .

@2above

Designerd wrote: You know that $\frac{a^2+b^2}{2} \geq ab$, right? So you just need to show that $\frac{a^2+b^2}{2} + 1 \geq a + b$. Try using some squares . We can also multiply both sides by 2 and we will get $a^2+b^2+2 \geq 2a+2b$ => $(a^2-2a+1) + (b^2 -2b +1) \geq 0$ => $ (a-1)^2 +(b-1)^2 \geq 0$ and that always true , so done

The same inequality here for real numbers.

Note that $$(a-b)^2 + (a-1)^2 + (b-1)^2 \ge 0 \Longleftrightarrow 2a^2 + 2b^2 -2ab-2a-2b+2 \ge 0 \Longleftrightarrow a^2 + b^2 -ab - a -b + 1 \ge 0 \Longleftrightarrow a^2 + b^2 + 2 \ge ab + a + b + 1 \Longleftrightarrow \sqrt{a^2 + b^2 + 2} \ge \sqrt{(a+1)(b+1)}$$Therefore, $$\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge \sqrt{(a+1)(b+1)} + \sqrt{(a+1)(c+1)} + \sqrt{(b+1)(c+1)} \ge 3 \sqrt[3]{(a+1)(b+1)(c+1)} \ge 3 \sqrt[3]{8} \ge 6$$as desired.

parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $a + b + c + ab + bc + ca + abc = 7$. Prove that $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } \ge 6$ . We claim that $a+b+c \ge 3$ Proof: Note that the given condition is equivalent to $(a+1)(b+1)(c+1) =8$. Hence by AM-GM, $$\frac{a+b+c+3}{3}=\frac{(a+1)+(b+1)+(c+1)}{3} \ge \sqrt[3]{(a+1)(b+1)(c+1)}=2$$Which implies our claim. Now note that $$\sum_{cyc} \sqrt{a^2+b^2+2} =\sum_{cyc} \sqrt{a^2+b^2+1^2+1^2} \overset{\text{C-S}}{\ge} \sum_{cyc} \frac{a+b+2}{2} =a+b+c+3\ge 6$$Which was to be proved. $\blacksquare$

It is easy to see that $\sqrt{x}$ is concave, so we can apply Jensen's inequality. After some manipulations, we get that the problem is equivalent to proving that $a^2+b^2+c^2\geq 3$. The original condition actually factors as $\prod (a+1)=8$, and with the following claim, we will be able to finish the problem. Claim: $a^2+b^2+2\geq (a+1)(b+1)$ Proof: This basically reduces to proving that $(a-1)(b-1)>0,$ which is trivially true for $a,b>0$. Now, we just have that $a^2+b^2+c^2\geq 3$ is equivalent to proving that $\sum_{cyc}(a^2+b^2+2)\geq 12,$ or that $\sum (a+1)(b+1)\geq 12,$ which follows from AM-MG.

Cute problem ! Lets use identities ! we know $(a+1)(b+1)(c+1) = a+b+c+ab+ac+bc+abc+1$ . but we know $a+b+c+ab+ac+bc+abc=7$. By adding $1$ to each side of equation we get $$(a+1)(b+1)(c+1)=8$$( ) So ! . Now we back to problem ! Let $\sqrt{a^2 + b^2 + 2 }+\sqrt{b^2 + c^2 + 2 }+\sqrt{c^2 + a^2 + 2 } = A$ By AM-GM inequality we get $\sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)} \geq 3\sqrt[3]{(a+1)(b+1)(c+1)}$ . ( ) By ( ) and ( ) we get $A\geq 6$ $\blacksquare$

TwinPrime wrote: I understand how we get $\sqrt{(a+1)(b+1)}+\sqrt{(b+1)(c+1)}+\sqrt{(c+1)(a+1)} \geq 3\sqrt[3]{(a+1)(b+1)(c+1)}$ from the AM-GM inequality, and how we get $3\sqrt[3]{(a+1)(b+1)(c+1)}=6$ based on what's given to us in the problem, but I don't really get how the solution above gets $a^2+b^2+2\geq(a+1)(b+1).$ It seems obvious, but I can't prove it. $a+b+c+ab+bc+ca+abc+1=(a+1)(b+1)(c+1)=8$

It is trivial that $$\sum{\sqrt{(a^2+1)+(b^2+1)}}\ge\sum{\sqrt{\frac{(a+1)^2+(b+1)^2}{2}}}$$From here, by $AM-GM$ $$\sum{\sqrt{\frac{(a+1)^2+(b+1)^2}{2}}} \ge\sum{\sqrt{(a+1)(b+1)}} \ge3\sqrt[6]{(a+1)^2(b+1)^2(c+1)^2}=6$$Which follows from the given statement.

Just Minkowski

Since $a+b+c+ab+bc+ca+abc+1=7+1=8=(a+1)(b+1)(c+1)$, let $a=2x-1$, $b=2y-1$, $c=2z-1$, therefore $xyz=1$. We want to prove that $$\sum_{cyc} \sqrt{4x^2-4x+4y^2-4y+4} \geq 6$$$$\iff \sum_{cyc} \sqrt{x^2-x+y^2-y+1} \geq 3$$But, we can use Cauchy-Schawrz to prove that $$4(x^2-x+y^2-y+1)=[(x-\frac{1}{2})^2+(y-\frac{1}{2})^2+\frac{1}{2}][1+1+2]\geq (x+y)^2$$. Hence $$\sum_{cyc} x^2-x+y^2-y+1 \geq \sum_{cyc} \frac{x+y}{2}=x+y+z$$. So, by AM-GM we´re done since $$x+y+z\geq 3\sqrt[3]{xyz}=3$$$\blacksquare$

We can factor the given equation to \[(a + 1)(b + 1)(c + 1) = 8\] Lemma 1: $a^2 + b^2 + 2 \geq (a+1)(b+1)$ Proof: We split up the inequality as follows: \[\frac{a^2 + b^2}{2} + \frac{a^2 + b^2}{2} + 1 \geq ab + a + b\] By AM-GM, we know that $\frac{a^2 + b^2}{2} \geq ab$. Thus, we just need to prove that \[\frac{a^2 + b^2}{2} + 1 \geq a + b\] Expanding and doing algebraic manipulations yields \[(a^2 - 2a + 1) + (b^2 - 2b + 1) \geq 0\]\[(a - 1)^2 + (b - 1)^2 \geq 0\] which is true by the trivial inequality. Thus, \[a^2 + b^2 + 2 \geq (a+1)(b+1)\] as desired. From Lemma 1, we know that $\sqrt{a^2 + b^2 + 2} \geq \sqrt{(a+1)(b+1)}$. We then need to prove that \[\sum_{\text{cyc}} \sqrt{(a+1)(b+1)} \geq 6\] By AM-GM, we have that \[\sum_{\text{cyc}} \sqrt{(a+1)(b+1)} \geq 3\sqrt[3]{\prod_{\text{cyc}} \sqrt{(a + 1)^2}}\]\[\sum_{\text{cyc}} \sqrt{(a+1)(b+1)} \geq 3\sqrt[3]{\prod_{\text{cyc}} a + 1}\] From the initial condition, we have that \[\sum_{\text{cyc}} \sqrt{(a+1)(b+1)} \geq 3\sqrt[3]{8}\]\[\sum_{\text{cyc}} \sqrt{(a+1)(b+1)} \geq 6\]\[\sum_{\text{cyc}} \sqrt{a^2 + b^2 + 2} \geq 6\] as desired.