Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$

Question

Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ : $$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$

I tried using AM-GM or AM-HM but I can't figure it out. $$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\root3\of{\Big(a+\frac{1}{b}\Big)^2\Big(b+\frac{1}{c}\Big)^2\Big(c+\frac{1}{a}\Big)^2}$$

$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(\Big(a+\frac{1}{b}\Big)\Big(b+\frac{1}{c}\Big)\Big(c+\frac{1}{a}\Big)\bigg)^{2/3}$$

$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(abc+a+b+c+\frac1a+\frac1b+\frac1c+\frac{1}{abc}\bigg)^{2/3}$$

$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(2+a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2/3}$$

Answer 1

By Cauchy we have $$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq \underbrace{{1\over 3}\bigg(a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2}}_B$$

By AM-GM and assumption $abc=1$ we have $$\frac1a+\frac1b+\frac1c \geq 3$$ so $$B \geq {1\over 3}\underbrace{\bigg(a+b+c+3\bigg)^{2}}_C$$

Let $x=a+b+c\geq 3$ then we have to check if $$C\geq 9(x+1)$$ or $$(x+3)^2\geq 9(x+1)\iff x(x-3)\geq 0$$

which is true.

Answer 2

Let us show stronger, with an intermediate step: $$ \begin{aligned} \left(a+\frac{1}{b}\right)^2 + \left(b+\frac{1}{c}\right)^2 + \left(c+\frac{1}{a}\right)^2 &\geq 4(a+b+c) \\ &\geq 3(a+b+c+1) \ . \end{aligned} $$ The last inequality, compactly written $a+b+c\ge3\sqrt[3]{abc}=3$ is simple, so let us show the first one. We write $$ a=A^3\ ,\qquad b=B^3\ ,\qquad c=C^3\ ,\qquad A,B,C>0\ ABC=1\ $$ and the given inequality, written first as $\sum(a+ac)^2\ge \dots$, becomes the homogeneous inequality: $$ A^6(C^3+ABC)^2+ B^6(A^3+ABC)^2+ C^6(B^3+ABC)^2 \ge 4(A^3+B^3+C^3)A^3B^3C^3\ . $$ At this point we can forget the constraint $ABC=1$. Let us see how to dominate with the terms on the left side the ones on the right side. First, let us expand $(C^3+ABC)^2 = C^6 + 2\,C^3\cdot ABC + (ABC)^2$. Then: $$ \begin{aligned} A^6C^6+B^6A^6+C^6B^6 &\ge (A^3+B^3+C^3)A^3B^3C^3\ , \\ &\qquad\text{which follows from} \\ &\qquad\frac 12A^6C^6+\frac 12B^6A^6\ge A^6B^3C^3 \\ &\qquad\text{and the cycled versions of it.} \\[2mm] 2(A^6C^3+B^6A^3+C^6B^3)ABC &\ge 2(A^3+B^3+C^3)A^3B^3C^3\ , \\ &\qquad\text{which follows from} \\ &\qquad\frac 23A^6C^3+\frac 13B^6A^3\ge (A^6C^3)^{2/3}(B^6A^3)^{1/3}=A^5B^2C^2 \\ &\qquad\text{and the cycled versions of it. Finally:} \\ (A^6+B^6+C^6)A^2B^2C^2 &\ge (A^3+B^3+C^3)A^3B^3C^3 \\ &\qquad\text{follows from} \\ &\qquad\frac 46A^6+\frac 16B^6+\frac 16C^6\ge A^4BC \\ &\qquad\text{and the cycled versions of it.} \end{aligned} $$ We have used the generalized form of the AM-GM inequality with more general weights $\lambda_1,\lambda_2,\dots,\lambda_n\ge 0$, total weight being one, $$\sum \lambda_kx_k\ge \prod x_k^{\lambda_k}\ .$$ (This is the Jensen inequality for the concave function $\ln$.)

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Note: The idea was simple, it is the standard way to show such inequalities. Typing all the stuff...

Answer 3

By AM-GM $$\sum_{cyc}\left(a+\frac{1}{b}\right)^2-3(a+b+c+1)=\sum_{cyc}\left(a^2+\frac{2a}{b}+\frac{1}{b^2}-3a-1\right)\geq$$ $$\geq\sum_{cyc}(2a-1+2+a^2b^2-3a-1)=\sum_{cyc}(a^2b^2-a)=$$ $$=\sum_{cyc}(a^2b^2-a^2bc)=\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$