Prove $2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a)$ for $abc=1.$

Question

Let $a,b,c>0: abc=1.$ Prove that: $$2(a+b+c)\left(1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge 3(a+b)(b+c)(c+a). $$

I've tried to use a well-known lemma but the rest is quite complicated for me.

Lemma. For any positive real numbers $x,y,z: xyz=1$ then $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\sqrt{3(x^2+y^2+z^2)}.$$ Proof for lemma.

Obviously, we need prove that $$\frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+2\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\right)\ge3(x^2+y^2+z^2).$$

By AM-GM $$\frac{x^2}{y^2}+\frac{x}{z}+\frac{x}{z}\ge3\sqrt[3]{\frac{x^4}{y^2z^2}}=3x^2,$$ $$\frac{y^2}{z^2}+\frac{y}{x}+\frac{y}{x}\ge3\sqrt[3]{\frac{y^4}{x^2z^2}}=3y^2,$$ and $$\frac{z^2}{x^2}+\frac{z}{y}+\frac{z}{y}\ge3\sqrt[3]{\frac{z^4}{y^2x^2}}=3z^2.$$ Thus, the lemma is proven. Equality holds iff $x=y=z=1.$

Back on the main problem, by applying the lemma, it's enough to prove $$2(a+b+c)\left(1+\sqrt{3(a^2+b^2+c^2)}\right)\ge 3(a+b)(b+c)(c+a). $$

I stopped here. It seems that we can square but it's quite complicated.

Hope you help me prove my last inequality. Thank you for your help.

Answer 1

Proof.

Firstly, rewrite the inequality as: $$2(a+b+c)(ab+bc+ca)(a^2c+c^2b+b^2a+1)\ge3(a+b)(b+c)(c+a)(ab+bc+ca)$$, $$\iff \frac{1+a^2c+c^2b+b^2a}{(a+b)(b+c)(c+a)}+1+a^2c+c^2b+b^2a\ge\frac{3(ab+bc+ca)}{2},$$ $$\iff 1+a^2c+c^2b+b^2a+\frac{a(a+b)(b+c)+b(b+c)(c+a)+c(c+a)(a+b)-(a+b)(b+c)(c+a)}{(a+b)(b+c)(c+a)}\ge\frac{3(ab+bc+ca)}{2},$$ $$\iff \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}\ge\frac{3(ab+bc+ca)}{2},$$ Or $$\frac{a}{b(a+c)}+\frac{b}{c(b+a)}+\frac{c}{a(c+b)}\ge \frac{3(ab+bc+ca)}{2(a+b+c)}. \tag{*}$$ From here, we can use the lemma you proved.

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Back on main problem, using C-S inequality and the lemma for (**): $$\frac{\dfrac{ac}{bc}}{a+c}+\frac{\dfrac{ba}{ca}}{a+b}+\frac{\dfrac{cb}{ab}}{b+c}\ge\frac{\left(\sqrt{\dfrac{ab}{bc}}+\sqrt{\dfrac{bc}{ca}}+\sqrt{\dfrac{ca}{ab}}\right)^2}{2(a+b+c)}\ge\frac{3(ab+bc+ca)}{2(a+b+c)}$$The problem is completely proven. Equality holds iff: $a=b=c=1.$

Answer 2

Symmetrization helps.

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

Thus,by AM-GM $x\geq1$ and we need to prove that $$u\left(2w^3+2\sum_{cyc}a^2c\right)\geq(9uv^2-w^3)w$$ or $$u\left(2w^3+\sum_{cyc}(a^2b+a^2c)\right)-(9uv^2-w^3)w\geq u\sum_{cyc}(a^2b-a^2c)$$ or $$(u-v)(9uv^2-w^3)\geq u(a-b)(a-c)(b-c)$$ and it's enough to prove that: $$(u-w)^2(9uv^2-w^3)^2\geq u^2\prod_{cyc}(a-b)^2$$ or $$(u-w)^2(9uv^2-w^3)^2\geq27u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $f(v^2)\geq0,$ where $$f(v^2)=108u^2v^6-81(2u-w)u^2wv^4-18(10u^2-2uw+w^2)uw^3v^2+108u^5w^3+28u^2w^6-2uw^7+w^8.$$

Now, $f'(v^2)=0$ gives $$v^2_{min}=\frac{3(2u-w)uw+w\sqrt{36u^4+44u^3w-7u^2w^2+8uw^3}}{12u}=$$ $$=\frac{3(2x-1)x+\sqrt{36x^4+44x^3-7x^2+8x}}{12x}\cdot w^2$$ and it remains to prove that: $f\left(v^2_{min}\right)\geq0$ or $$648x^5-396x^4+342x^3+107x^2+20x+8\geq\sqrt{(36x^3+44x^2-7x+8)^3x}$$ or $$9x(648x^5-396x^4+342x^3+107x^2+20x+8)\geq\sqrt{81x^3(36x^3+44x^2-7x+8)^3}.$$ But by AM-GM $$\sqrt{81x^3(36x^3+44x^2-7x+8)^3}\leq\left(\frac{81x^3+3(36x^3+44x^2-7x+8)}{4}\right)^2.$$ Id est, it's enough to prove that: $$9x(648x^5-396x^4+342x^3+107x^2+20x+8)\geq\left(\frac{81x^3+3(36x^3+44x^2-7x+8)}{4}\right)^2$$ or $$(x-1)^2(6399x^4+918x^3-145x^2+112x-64)\geq0$$ and we are done!

Answer 3

Some Remarks: SOS-Schur method and Buffalo Way (BW)

Using the substitution $a = \frac{y}{x}, b = \frac{z}{y}, c = \frac{x}{z}$, it suffices to prove that, for all $x, y, z > 0$, \begin{align*} \sum_{\mathrm{cyc}} \left(2\,{x}^{5}y-3\,{x}^{4}yz-3\,{x}^{3}{y}^{3}+4\,{x}^{3}{y}^{2}z+2\,{x}^{ 2}{y}^{4}-2\,{x}^{2}{y}^{2}{z}^{2}\right) \ge 0. \tag{1} \end{align*}

• SOS-Schur method

See: here, here.

WLOG, assume that $z = \max(x, y, z)$. (1) can be written as the form $$f(x, y, z) = M(x - y)^2 + N(z - x)(z - y)$$ where $M, N \ge 0$.

There are many such $M, N$.

For example, Andreas' method: \begin{align*} M &= {x}^{3}y+{x}^{3}z+2\,{x}^{2}{y}^{2}+{x}^{2}yz+x{y}^{3}+x{y}^{2}z+{y}^{ 3}z , \\ N &= {x}^{4}-{x}^{3}z-{x}^{2}{y}^{2}+3\,{x}^{2}yz+2\,{x}^{2}{z}^{2}+2\,x{y} ^{3}+x{y}^{2}z\\ &\qquad -xy{z}^{2}+2\,x{z}^{3}-{y}^{4}-{y}^{3}z+2\,{y}^{2}{z}^{2 }. \end{align*} (Note: It is not difficult to prove that $N \ge 0$.)

Using computer, we can obtain the same $M, N$ as Andreas' one. We can also obtain the following: \begin{align*} M &= 2\,{x}^{3}y+3\,{x}^{2}{y}^{2}-{x}^{2}yz+{x}^{2}{z}^{2}+x{y}^{3}+xy{z}^ {2}+{y}^{3}z, \\ N &= {x}^{3}y-{x}^{3}z+3\,{x}^{2}yz+2\,{x}^{2}{z}^{2}+x{y}^{3}+x{y}^{2}z-xy {z}^{2}+2\,x{z}^{3}\\ &\qquad -{y}^{4}-{y}^{3}z+2\,{y}^{2}{z}^{2}. \end{align*} (Note: It is easy to prove that $M, N\ge 0$.)

• BW (Buffalo Way):

WLOG, assume that $z = \max(x, y, z)$. Let $x = \frac{z}{1 + s}, y = \frac{z}{1 + t}$ for $s, t \ge 0$. (1) is equivalently written as \begin{align*} &2\,{s}^{4}{t}^{5}+2\,{s}^{5}{t}^{3}+7\,{s}^{4}{t}^{4}+8\,{s}^{3}{t}^{5 }+3\,{s}^{5}{t}^{2}+18\,{s}^{4}{t}^{3}+32\,{s}^{3}{t}^{4}\\ &\qquad +9\,{s}^{2}{t }^{5}+21\,{s}^{4}{t}^{2}+62\,{s}^{3}{t}^{3}+39\,{s}^{2}{t}^{4}+4\,s{t} ^{5}+{s}^{5}+3\,{s}^{4}t\\ &\qquad +60\,{s}^{3}{t}^{2}+72\,{s}^{2}{t}^{3}+17\,s{t }^{4}+{t}^{5}+5\,{s}^{4}+12\,{s}^{3}t+63\,{s}^{2}{t}^{2}\\ &\qquad +28\,s{t}^{3}+ 5\,{t}^{4}+10\,{s}^{3}+9\,{s}^{2}t+17\,s{t}^{2}+10\,{t}^{3}+8\,{s}^{2} -8\,st+8\,{t}^{2}\\ &\ge 0 \end{align*} which is clearly true.