Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $

Question

I tried this question in two ways-

Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$

Approach 1: $$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$ $$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$ $$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$

I reached till here but can't take it forward.

Approach 2: $$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$ but it failed as $$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$ This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!

Answer 1

Another way.

By C-S and AM-GM we obtain: $$\prod_{cyc}(1+a^2)\geq(a+b)^2(1+c^2)=(a+b)^2+(ac+bc)^2\geq$$ $$\geq4ab+(3-ab)^2=a^2b^2-2ab+9=(ab-1)^2+8\geq8.$$

Answer 2

Let $a=\sqrt3\tan\alpha$, $b=\sqrt3\tan\beta$ and $c=\sqrt3\tan\gamma,$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right)$.

Thus, $$\alpha+\beta+\gamma=\frac{\pi}{2}$$ and we need to prove that: $$\ln\left((1+a^2)(1+b^2)(1+c^2)\right)\geq3\ln2$$ or $$\ln(1+a^2)-\ln2+\ln(1+b^2)-\ln2+\ln(1+c^2)-\ln2\geq0$$ or $$\sum_{cyc}(\ln(1+a^2)-\ln2)\geq0$$ or $$\sum_{cyc}\left(\ln(1+3\tan^2\alpha\right)-\ln2)\geq0$$ or $$\sum_{cyc}\left(\ln\left(1+3\tan^2\alpha\right)-\ln2-\frac{4}{\sqrt3}\left(\alpha-\frac{\pi}{6}\right)\right)\geq0,$$ which is true because easy to show that for any $x\in\left(0,\frac{\pi}{2}\right)$ we have $f(x)\geq0,$ where $$f(x)=\ln\left(1+3\tan^2x\right)-\ln2-\frac{4}{\sqrt3}\left(x-\frac{\pi}{6}\right).$$

Also, we see that $$f''(x)=\frac{6(4\cos^4x-6\cos^2x+3)}{\cos^2x(2-\cos2x)^2}>0$$ and our inequality follows from Jensen.

Answer 3

Another way.

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

Thus, $v^2=1$ and we need to prove that: $$a^2b^2c^2+\sum_{cyc}(a^2b^2+a^2)\geq7$$ or $f(w^3)\geq0,$ where $$f(w^3)=w^6+(9v^4-6uw^3)v^2+(9u^2-6v^2)v^4-7v^6.$$ But, $$f'(w^3)=2w^3-6uv^2w^3<0,$$ which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,

which by $uvw$ happens for equality case of two variables.

Since our inequality is symmetric, we can assume that $b=a$, which gives $c=\frac{3-a^2}{2a}.$

Can you end it now?

For $a^2=x$ I got $$(x-1)^2(x^2+2x+9)\geq0.$$

Answer 4

Another way.

We need to prove that $$\prod_{cyc}(3+3a^2)\geq8\cdot27$$ or $$\prod_{cyc}(ab+ac+bc+3a^2)\geq8(ab+ac+bc)^3$$ or $$\sum_{sym}(3a^4b^2+a^3b^3+3a^4bc-6a^3b^2c-a^2b^2c^2)\geq0,$$ which is true by Muirhead because

$(4,2,0\succ(3,2,1),$ $(4,1,1)\succ(3,2,1)$ and $(3,3,0)\succ(2,2,2).$

Answer 5

Another way.

Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain: $$(1+a^2)(1+b^2)(1+c^2)=((1-ab)^2+(a+b)^2)(1+c^2)=$$ $$=(1-ab-(a+b)c)^2+((1-ab)c+a+b)^2=$$ $$=(1-ab-ac-bc)^2+(a+b+c-abc)^2=4+(a+b+c-abc)^2.$$ Id est, it's enough to prove that $$a+b+c-abc\geq2,$$ which is true by Maclaurin.

Indeed, let $a+b+c=3u$, $ab+ac+bc=v^2$, where $v>0$ and $abc=w^3$.

Thus, we need to prove that $$3u-w^3\geq2$$ or $$3u^2v-2v^3\geq w^3,$$ which is true because $$u\geq v\geq w.$$

Answer 6

Here is another solution using AM-GM and Cauchy-Schwarz. Our goal is to prove that: $$(1+a^2)(1+b^2)(1+c^2)\geq8$$

Rewriting the L.H.S. expression a little, we have: $$(a^2+1)(1+b^2)(1+c^2) \ge (a+b)^2(1+c^2)=S,$$ where we have applied C-S to the first two terms in the product.

Hence it suffices for us to show that: $$S=a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge (ab+bc+ca)^2-1$$ $$\iff a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge a^2b^2+2a^2bc+a^2c^2+2ab^2c+2abc^2+b^2c^2-1$$ $$\iff a^2+2ab+b^2 \ge a^2b^2+2a^2bc+2ab^2c-1$$ $$\iff (a+b)^2+1 \ge ab(ab+2ac+2bc)$$ $$\iff (a+b)^2+1 \ge ab(3+ac+bc) $$ $$\iff a^2+b^2+1 \ge ab(1+ac+bc)=ab(4-ab)$$ $$\iff a^2+b^2+a^2b^2+1 \ge 4ab$$

But the last inequality is trivially true by AM-GM; since: $$a^2+b^2+a^2b^2+1 \ge 4 \sqrt[^4]{a^4b^4}=4ab,$$ and we are done.

Answer 7

the pqr method:

Let $p = a+b+c, q = ab+bc+ca = 3, r = abc$.
Since $(a+b+c)^2 - 3(ab+bc+ca) = a^2+b^2+c^2 - ab-bc-ca \ge 0$, we have $p \ge 3$.
Since $ab+bc+ca \ge 3\sqrt[3]{ab \cdot bc \cdot ca} = 3\sqrt[3]{(abc)^2}$, we have $r\le 1$.

We have \begin{align} (1+a^2)(1+b^2)(1+c^2) - 8 &= a^2b^2c^2+a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2-7\\ &= r^2 + q^2 - 2pr + p^2-2q - 7 \\ &= (p-r)^2 - 4 \\ &\ge (3-1)^2 - 4\\ &= 0. \end{align} We are done.