# Prove/disprove $\sum_{cyc}a\sqrt{\frac{(ca + 1)(ab + 1)}{bc + 1}} \ge 2$ where $a$, $b$, $c > 0$ and $a^2 + b^2 + c^2 = 1$

$$a$$, $$b$$ and $$c$$ are positives such that $$a^2 + b^2 + c^2 = 1$$. Prove/disprove that $$a\sqrt{\frac{(ca + 1)(ab + 1)}{bc + 1}} + b\sqrt{\frac{(ab + 1)(bc + 1)}{ca + 1}} + c\sqrt{\frac{(bc + 1)(ca + 1)}{ab + 1}} \ge 2$$

I tried using the Bunyakovsky inequality for $$(a, b, c)$$ and $$\left(\sqrt{\frac{(ca + 1)(ab + 1)}{bc + 1}}, \sqrt{\frac{(ab + 1)(bc + 1)}{ca + 1}}, \sqrt{\frac{(bc + 1)(ca + 1)}{ab + 1}}\right)$$ but realized it is the opposite inequality sign.

If the inequality is not correct, what minor change could have been done to the inequality so that it is correct for $$\forall a, b, c \in \mathbb R^+|a^2 + b^2 + c^2 = 1$$?

The second part of the problem is quite subjective, so check this problem out as an example.

• consider $a=1,b=0,c=0$ – Bonbon May 1 at 15:12
• $a$, $b$ and $c$ are positives, so $a$, $b$, $c > 0$. – Lê Thành Đạt May 1 at 15:15
• I mean but a,b,c, can be sufficiently close to this set of value, so the inequality is wrong – Bonbon May 1 at 15:18
• Well, can you answer the second part of the question? – Lê Thành Đạt May 1 at 15:19
• Maybe change into $\le 2$? I am thinking about it. – Bonbon May 1 at 15:21

It's wrong. Try $$b=c\rightarrow0^+$$.

The following inequality is true already.

Let $$a$$, $$b$$ and $$c$$ be non-negative numbers such that $$a^2+b^2+c^2=1$$. Prove that: $$\sum_{cyc}a\sqrt{\frac{(ab+1)(ac+1)}{bc+1}}\leq2.$$

Indeed, let $$a+b+c=3u$$, $$ab+ac+bc=3v^2$$ and $$abc=w^3$$.

Thus, $$9u^2-6v^2=1$$ and we need to prove that: $$\sum_{cyc}a(ab+1)(ac+1)\leq2\sqrt{\prod_{cyc}(ab+1)}$$ or $$\sum_{cyc}(a^3bc+a^2b+a^2c+a)\leq2\sqrt{a^2b^2c^2+1+\sum_{cyc}(a^2bc+ab)}$$ or $$abc+\sum_{cyc}(a^2b+a^2c+a)\leq2\sqrt{a^2b^2c^2+1+\sum_{cyc}(a^2bc+ab)}$$ or $$9uv^2-2w^3+3u(9u^2-6v^2)\leq$$ $$\leq2\sqrt{w^6+(9u^2-6v^2)^3+3uw^3(9u^2-6v^2)+3v^2(9u^2-6v^2)^2}.$$ Now, let $$f(w^3)=2\sqrt{w^6+(9u^2-6v^2)^3+3uw^3(9u^2-6v^2)+3v^2(9u^2-6v^2)^2}-$$ $$-(9uv^2-2w^3+3u(9u^2-6v^2)).$$ Thus, it's obvious that $$f$$ increases.

Id est, it's enough to prove our inequality for an extreme value of $$w^3$$, which happens in the following cases.

1. $$w^3=0$$.

The homogenization gives $$abc+\sum_{cyc}(a^2b+a^2c)+(a+b+c)(a^2+b^2+c^2)\leq2\sqrt{\prod_{cyc}(ab+a^2+b^2+c^2)}.$$ Since this inequality is an even degree and for $$b=c=0$$ it's obviously true, it's enough to assume $$b=1$$ and $$c=0$$, which gives $$a^2+a+(a+1)(a^2+1)\leq2(a^2+a+1)\sqrt{a^2+1}$$ or $$a+1\leq2\sqrt{a^2+1},$$ which is true because by C-S $$2\sqrt{a^2+1}\geq\sqrt{(1+1)(a^2+1)}\geq a+1.$$ 2. Two variables are equal.

We can assume $$b=c=1$$ and it's enough to prove that: $$a+2(a^2+a+1)+(a+2)(a^2+2)\leq2(a^2+a+2)\sqrt{a^2+3}$$ or $$2\sqrt{a^2+3}\geq a+3,$$ which is true by C-S again: $$2\sqrt{a^2+3}=\sqrt{(1+3)(a^2+3)}\geq a+3$$ and we are done!

Also, we can prove that $$f(w^3)\geq0$$ by the following way.

We need to prove that: $$2\sqrt{w^6+(9u^2-6v^2)^3+3uw^3(9u^2-6v^2)+3v^2(9u^2-6v^2)^2}\geq$$ $$\geq9uv^2-2w^3+3u(9u^2-6v^2)$$ or $$2\sqrt{w^6+(9u^2-6v^2)^2(9u^2-3v^2)+3uw^3(9u^2-6v^2)}\geq27u^3-9uv^2-2w^3.$$ Now, by AM-GM and C-S we obtain: $$2\sqrt{w^6+(9u^2-6v^2)^2(9u^2-3v^2)+3uw^3(9u^2-6v^2)}\geq$$ $$\geq2\sqrt{w^6+6u^2(9u^2-6v^2)^2+9w^6}=2\sqrt{10w^6+54u^2(3u^2-2v^2)^2}=$$ $$=\frac{1}{4}\sqrt{(10+54)(10w^6+54u^2(3u^2-2v^2)^2)}\geq\frac{1}{4}(10w^3+54u(3u^2-2v^2))=$$ $$=\frac{1}{2}(5w^3+81u^3-54uv^2).$$ Id est, it's enough to prove that: $$\frac{1}{2}(5w^3+81u^3-54uv^2)\geq27u^3-9uv^2-2w^3$$ or $$3u^3-4uv^2+w^3\geq0,$$ which is Schur.

• It is such a shame that I am still a 9th-grade student and I haven't studied derivatives yet. Could you post a different answer please, you needn't delete your current one. – Lê Thành Đạt May 21 at 10:02
• @Lê Thành Đạt We can see that $f$ increases without derivatives. – Michael Rozenberg May 21 at 10:36
• @Lê Thành Đạt I added another way. – Michael Rozenberg May 21 at 11:10
• Thanks for that so much! – Lê Thành Đạt May 21 at 13:50

Claim: For nonnegative real number a,b,c such that $$\sum_{cyc}a^2=1$$, we have $$\sum_{cyc}a\sqrt{\frac{(ab+1)(ac+1)}{bc+1}}\le 2.$$

Note that by multipling $$\sqrt{(ab+1)(ac+1)(bc+1)}$$ on both sides, and taking the square, the inequality is equivalent to $$\Delta=4(ab+1)(ac+1)(bc+1)-(\sum_{cyc}a(ab+1)(ac+1))^2\ge 0.$$ Homogenizing $$\Delta$$, we get $$\Delta=4(\sum_{cyc}a^2)^2(ab+\sum_{cyc}a^2)(ac+\sum_{cyc}a^2)(bc+\sum_{cyc}a^2)-(\sum_{cyc}a(ab+\sum_{cyc}a^2)(ac+\sum_{cyc}a^2))$$, it is homogeneous of degree 10 and symmetry for $$a,b,c$$.

Now, assume $$a=min(a,b,c)$$ and $$b=a+s$$,$$c=a+t$$, then $$s,t\ge0$$. Taking the substitution, with the help of calculator, we have

$$\Delta=432(t^2−st+s^2)a^8+(1224t^3−108st^2−108s^2t+1224s^3)a^7+(1803t^4+678st^3+369s^2t^2+678s^3t+1803s^4)a^6+(1686t^5+1194st^4+930s^2t^3+930s^3t^2+1194s^4t+1686s^5)a^5+(1089t^6+948st^5+1164s^2t^4+702s^3t^3+1164s^4t^2+948s^5t+1089s^6)a^4+(492t^7+456st^6+732s^2t^5+528s^3t^4+528s^4t^3+732s^5t^2+456s^6t+492s^7)a^3+(153t^8+126st^7+301s^2t^6+174s^3t^5+280s^4t^4+174s^5t^3+301s^6t^2+126s^7t+153s^8)a^2+(30t^9+18st^8+70s^2t^7+38s^3t^6+60s^4t^5+60s^5t^4+38s^6t^3+70s^7t^2+18s^8t+30s^9)a+(3t^{10}+10s^2t^8−2s^3t^7+15s^4t^6−4s^5t^5+15s^6t^4−2s^7t^3+10s^8t^2+3s^{10})\ge0$$, hence the claim is true.

• But could you prove the inequality with classical inequalities? – Lê Thành Đạt May 2 at 11:38
• @LêThànhĐạt Well, maybe, but I have no good idea. The $bc+1$ on the denominator stuck me. – Bonbon May 2 at 15:02
• Well, maybe the Bunyakovsky inequality can help, I am not quite sure though. – Lê Thành Đạt May 2 at 15:03