Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$

Question

The question is this:

If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then prove that $$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$$

For my work on this inequality, I have proved already under constraints that it is true.

Proof for: $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0.$ $$ \sqrt{3}abc + \sqrt{2}a - \sqrt{3} - \sqrt{2}c \geqslant 0 $$ $$ a\left( \sqrt{3}bc + \sqrt{2} \right) + (-1)\left( \sqrt{3} + \sqrt{2}c \right) \geqslant 0 $$ $$ (1 + 1)(a\left( \sqrt{3}bc + \sqrt{2} \right) + (-1)\left( \sqrt{3} + \sqrt{2}c \right)) \geqslant 0 $$ By Chebyshev, $$ (a - 1) (\sqrt{3}bc + \sqrt{2} + \sqrt{3} + \sqrt{2}c )\geqslant0 $$ $$ a \geqslant 1 $$ Chebyshev Inequality requires the sequences to be monotonous. As $a+1>0$, we need to have the other sequence also in the same order, hence the condition: $\sqrt{3}bc + \sqrt{2} \geqslant\sqrt{3} + \sqrt{2}c$. The sequences are $(a,-1)$ and $(\sqrt{3}bc + \sqrt{2} ,\sqrt{3} + \sqrt{2}c)$.

I have tried another way but that was untrue. I have reached this far. The constraint $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0$ isn't true always. Try $(a,b,c) = (\sqrt{3},0,0)$.

Thanks for extensions or other solutions too are welcome!

Answer 1

Denote \begin{align} P &= abc - 1 + \sqrt{\frac{2}{3}}(a-c),\\ Q &= \frac{a^2+b^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{\tfrac{a^2+b^2}{2}} - c). \end{align}

First, it is easy to prove $Q= \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{\tfrac{3-c^2}{2}} - c) \ge 0$ (note: $c\in [0,1]$). Indeed, if $c\in [0, \frac{1}{2}]$, we have \begin{align} Q &= \frac{3-c^2}{2}c - 1 + \sqrt{1 - \frac{c^2}{3}} - \sqrt{\tfrac{2}{3}}\ c\\ &\ge \frac{3-c^2}{2}c - 1 + 1 - \frac{c^2}{3} - \sqrt{\tfrac{2}{3}}\ c \\ &= \frac{1}{6}c(-3c^2 - 2c + 9 - 2\sqrt{6})\\ &\ge 0, \end{align} and if $c\in (\frac{1}{2}, 1]$, we have \begin{align} Q &= \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(\sqrt{1 + \tfrac{1-c^2}{2}} - c)\\ &\ge \frac{3-c^2}{2}c - 1 + \sqrt{\tfrac{2}{3}}(1 + \tfrac{1}{3}\cdot \tfrac{1-c^2}{2} - c)\\ &= \frac{1}{18}(1-c)[9c^2 + (\sqrt{6} + 9)c + 7\sqrt{6} - 18]\\ &\ge 0. \end{align}

Second, we have (let $x = \frac{b}{a} \in [0, 1]$) \begin{align} &P - Q\\ =\ & (ab - \tfrac{a^2+b^2}{2})c + \sqrt{\tfrac{2}{3}}(a - \sqrt{\tfrac{a^2+b^2}{2}})\\ =\ & \sqrt{\tfrac{2}{3}}\frac{\frac{a^2 - b^2}{2}}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - \frac{(a-b)^2}{2} c\\ =\ & \frac{a-b}{2} \left[\sqrt{\tfrac{2}{3}}\frac{a + b}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - (a-b)c\right]\\ \ge\ & \frac{a-b}{2} \left[\sqrt{\tfrac{2}{3}}\frac{a + b}{a + \sqrt{\tfrac{a^2+b^2}{2}}} - (a-b)b\right]\\ =\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + \sqrt{\tfrac{1+x^2}{2}}} - (1-x)x a^2\right]\\ \ge\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + \sqrt{\tfrac{1+x^2}{2}}} - (1-x)x \frac{3}{1+x^2}\right]\\ \ge\ & \frac{a-b}{2}\left[\sqrt{\tfrac{2}{3}}\frac{1 + x}{1 + 1 - \frac{1-x^2}{4}} - (1-x)x \frac{3}{1+x^2}\right]\\ \ge\ & \frac{a-b}{2}\cdot \frac{9x^4 + (4\sqrt{6}-9)x^3 + (4\sqrt{6}+63)x^2 + (4\sqrt{6}-63)x+4\sqrt{6}}{3(x^2+7)(x^2+1)}\\ \ge\ & \frac{a-b}{2}\cdot \frac{(4\sqrt{6}+63)x^2 + (4\sqrt{6}-63)x+4\sqrt{6}}{3(x^2+7)(x^2+1)}\\ \ge\ & 0 \end{align} where we have used $3 = a^2 + b^2 + c^2 \ge a^2 + a^2x^2$ to obtain $a^2 \le \frac{3}{1+x^2}$, and we have used $\sqrt{\tfrac{1+x^2}{2}} = \sqrt{1 - \frac{1-x^2}{2}} \le 1 - \frac{1-x^2}{4} $.

We are done.

Answer 2

Let $$f(a,b,c,\lambda)=abc-1+\sqrt{\frac{2}{3}}(a-c)+\lambda(a^2+b^2+c^2-3).$$ Thus, in the minimum point we need $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial\lambda}=0,$$ which gives: $$bc+\sqrt{\frac{2}{3}}+2\lambda a=ac+2\lambda b=ab-\sqrt{\frac{2}{3}}+2\lambda c=0.$$ Now, if $c=0$, so $$3=a^2+b^2\leq2a^2,$$ which gives $$a\geq\sqrt{\frac{3}{2}}$$ and $$abc-1+\sqrt{\frac{2}{3}}(a-c)=\sqrt{\frac{2}{3}}a-1\geq0.$$

Now, let $c>0$.

Thus, $$ \frac{bc+\sqrt{\frac{2}{3}}}{a}=\frac{ab-\sqrt{\frac{2}{3}}}{c}=\frac{ac}{b},$$ which gives $$b^2c+b\sqrt{\frac{2}{3}}=a^2c$$ and $$b^2a-b\sqrt{\frac{2}{3}}=ac^2,$$ which after summing gives $$b^2=ac,$$ which with our condition gives $$a^2+ac+c^2=3$$ and we need to prove that $$\sqrt{a^3c^3}+\sqrt{\frac{2}{3}}(a-c)\cdot\frac{a^2+ac+c^2}{3}\geq\sqrt{\left(\frac{a^2+ac+c^2}{3}\right)^3}.$$ Now, let $a=xc$ and $a^2+c^2=2uac.$

Thus, $x\geq1$ and $u\geq1$ and we need to prove that: $$\sqrt{x^3}+\sqrt{\frac{2(x^2+1-2x)}{3}}\cdot\frac{x^2+x+1}{3}\geq\sqrt{\left(\frac{x^2+x+1}{3}\right)^3}$$ or $$1+\sqrt{\frac{4(u-1)}{3}}\cdot\frac{2u+1}{3}\geq\sqrt{\left(\frac{2u+1}{3}\right)^3}$$ or $$27+4(u-1)(2u+1)^2+12\sqrt{3(u-1)}(2u+1)\geq(2u+1)^3$$ or $$(u-1)(4u^2-2u-11)+6\sqrt{3(u-1)}(2u+1)\geq0,$$ which is obvious for $4u^2-2u-11>0$ or $u>\frac{1+\sqrt{45}}{4}.$

Id est, it's enough to prove that $$6\sqrt{3}(2u+1)\geq\sqrt{u-1}(-4u^2+2u+11)$$ for $$1\leq u\leq \frac{1+\sqrt{45}}{4}.$$ Indeed, we need to prove that: $$108(2u+1)^2\geq(u-1)(4u^2-2u-11)^2$$ or $$229+355u+304u^2+68u^3+32u^4-16u^5\geq0$$ or $$229+355u+304u^2+24u^3+24u^4+4u^3(11+2u-4u^2)\geq0$$ and we are done in this case.

Also, we need to check, what happens for $b=c$ and for $a=b$.

Two these cases lead to inequalities of one variable.

I hope there is a solution without LM.

Answer 3

Let $f(a,b)=abc+(a-c)k$ where $c^2=3-a^2-b^2$ and $k=\sqrt{2/3}$. Assuming $c\ne0$, $$f_a=bc+(1-c_a)k=0$$ for critical points. Now $c_a=-a/c$ so $bc^2+(c+a)k=0$, contradicting $c\ne0$. This means that either $c=0$, or any other solutions must lie on the boundaries of the constraints, which are:

1. $a=b$ which yields $f(a)=(a^2-k)\sqrt{3-2a^2}+ak$;

2. $b=c$ which yields $f(a)=a(3-a^2)/2+(a-\sqrt{(3-a^2)/2})k$.

When $c=0$ we have $a^2+b^2=3$ such that $a\ge b$ so $a\ge\sqrt{3/2}$ and $f(a,b)=0+ak\ge1$.

For the first case we have $a\ge c\implies a\ge1$ so the domain of $f$ is $[1,\sqrt{3/2}]$. Notice that $f(1)=f(\sqrt{3/2})=1$ and $f(a)-1$ is positive. Similarly, for the second case we also have $a\ge1$. Note that $f(1)=1$ and calculus yields $f(a)\ge1$.

Answer 4

If $a\geq b\geq c\geq0$ then prove $$3\sqrt3abc+\sqrt2\left(a-c\right)\left(a^2+b^2+c^2\right)\geq\left(a^2+b^2+c^2\right)^{\frac{3}2}.$$ Case 1: $c=0,$ it's obvious. Equality at $a=b\iff a=b=\sqrt{\frac{3}2}.$

Case 2: $c=1.$ If $a=1,$ then we are done. Equality at $a=b=c=1.$ If $a>1$ then consider on $[1,a]$ the function $$f(b):=3\sqrt3ab+\sqrt2\left(a-1\right)\left(a^2+b^2+1\right)-\left(a^2+b^2+1\right)^{\frac{3}2}.$$ We have: $$f'(b)=b\left(\frac{3\sqrt3a}b+2\sqrt2\left(a-1\right)-3\sqrt{a^2+b^2+1}\right)\implies$$ $f$ is pseudo-concave $\implies\min_{b\in[1,a]}{f(b)}\in\{f(1),f(a)\}.$ But $$f(1)>0$$ and $$f(a)>\sqrt3\left(2a^2+1\right)+\sqrt2\left(a-1\right)\left(2a^2+1\right)-\left(2a^2+1\right)^{\frac{3}2}>0.$$ We are done. Edit: Let me give further details about $f(1)>0.$ We need to prove $$3\sqrt3a+\sqrt2\left(a-1\right)\left(a^2+2\right)>\left(a^2+2\right)^{\frac{3}2}\iff$$ $$6\sqrt6a\left(a-1\right)\left(a^2+2\right)>a\left(a-1\right)^2\left(-a^3+2a^2+a+16\right)\iff$$ $$a^4-3a^3+3a^2+a^2\left(-2+6\sqrt6\right)-15a+16+12\sqrt6>0,$$ which is obviously true.