Let a,b,c be positive real numbers numbers such that $ a^2 + b^2 + c^2 = 3 $

Question

Let $a,b,c\in\mathbb{R^+}$ such that $ a^2 + b^2 + c^2 = 3 $. Prove that $$ (a+b+c)(a/b + b/c + c/a) \geq 9. $$

My Attempt I tried AM-GM on the symmetric expression so the $a+b+c \geq 3$, but I found $a+b+c \leq 3$.

Answer 1

From Cauchy-Schwarz Inequality: $$\sum\limits_{cyc} \frac{a}{b}\ge \frac{(a+b+c)^2}{ab+bc+ca}$$

Hence, it suffices to prove: $\displaystyle \frac{(a+b+c)^3}{ab+bc+ca} \ge 9$

Squaring both sides and using $a^2+b^2+c^2 = 3$,

$$\begin{align} & (a+b+c)^6 \ge 81(ab+bc+ca)^2 = 27(a^2+b^2+c^2)(ab+bc+ca)^2 \end{align}$$

which follows from the Am-Gm Inequality with $3$ terms, $$a^2+b^2+c^2,ab+bc+ca,ab+bc+ca$$

as,

$\displaystyle \frac{(a+b+c)^6}{27} = \left(\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3}\right)^3 \ge (a^2+b^2+c^2)(ab+bc+ca)^2$

Answer 2

The homogeneous form of the inequality is $$(a+b+c)^2(ca^2+ab^2+bc^2)^2\geq 27(a^2+b^2+c^2)\,a^2 b^2 c^2 $$ that is equivalent to:

$$ (a^2+b^2+c^2+2ab+2bc+2ac)(c^2 a^4+a^2 b^4 + b^2 c^4 + 2 a^3 b^2 c+2 b^3 c^2 a+2 c^3 a^2 b) \geq 27(a^4 b^2 c^2+a^2 b^4 c^2 + a^2 b^2 c^4) $$ that can be proved by combining Schur's inequality with Muirhead's inequality.

Answer 3

Consider the following optimization problem $$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}$$ subject to $$x^2+y^2+z^2=3$$ All we need to show is that $$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=9$$ under the condition $$x^2+y^2+z^2=3$$ Set up the Lagrangian $$\mathcal{L}(x,y,z,\lambda)=(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+\lambda(3-x^2-y^2-z^2)$$ where $\lambda>0$. First order conditions wrt $x,y,z$ yield $$\mathcal{L}_x=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{y}-\frac{z}{x^2})=2\lambda x$$ $$\mathcal{L}_y=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{z}-\frac{x}{y^2})=2\lambda y$$ $$\mathcal{L}_z=0\Leftrightarrow(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})+(x+y+z)(\frac{1}{x}-\frac{y}{z^2})=2\lambda z$$ Exploiting the symmetry of the system of simultaneous equations above one obtains $$x^*=y^*=z^*$$ where $(x^*,y^*,z^*)$ is the solution to the minimization problem. Invoking here the constraint we get $$(x^*)^2+(y^*)^2+(z^*)^2=9\Rightarrow 3(x^*)^2=9\Rightarrow x^*=1$$ ($x^*=1$,since $x^*\geq0$). So $(x^*,y^*,z^*)=(1,1,1)$ minimizes the objective function i.e. $$\min_{x,y,z}\{(x+y+z)(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\}=(x^*+y^*+z^*)(\frac{x^*}{y^*}+\frac{y^*}{z^*}+\frac{z^*}{x^*})=(1+1+1)(\frac{1}{1}+\frac{1}{1}+\frac{1}{1})=9$$

Answer 4

We'll prove that $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\geq\frac{3}{2}\sum\limits_{cyc}\left(\frac{a}{b}+\frac{a}{c}\right)$.

Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.

Thus, we need to prove that $2(x+y+z)^2\geq3(x+y+z+xy+xz+yz)$,

where $x>0$, $y>0$ and $z>0$ such that $xyz=1$

or $\sum\limits_{cyc}(2x^2-3x+xy)\geq0$ or $\sum\limits_{cyc}\frac{(2x+1)(x-1)^2}{x}\geq0$, which is obvious.

Hence, it remains to prove that $(a+b+c)^2\sum\limits_{cyc}(a^2b+a^2c)\geq18abc(a^2+b^2+c^2)$.

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

Hence, we need to prove that $9u^2(9uv^2-3w^3)\geq18w^3(9u^2-6v^2)$ or $(7u^2-4v^2)w^3\leq3u^3v^2$.

But $w^3$ gets a maximal value, when two numbers from $\{a,b,c\}$ are equal.

Since the last inequality is homogeneous, it remains to check one case only: $b=c=1$,

which gives $(a-1)^2(a^2-2a+4)\geq0$. Done!