# An Inequality with a Parameter and a Constraint

### Solution

$\mathbf{\text{Case }\lambda=26.}$

We first reformulate the problem:

If $a,b,c\ge 0\,$ and $a+b+c=3,\,$ then $a^4+b^4+c^4+78abc\le 81.$

Indeed, let $a^2+b^2+c^2=3(1+2t^2),\,$ $0\le t\le 1\,$ and $abc=p.\,$ Then $a^4+b^4+c^4=3(6t^4+24t^2+4p-3).\,$ Suffice it to prove that $6t^4+24t^2-30+30p\le 0.\,$

But $p\le 2t^3-3t^2+1,\,$ hence, suffice it to prove that $2t^4+8t^2-10+10(2t^3-3t^2+1)\le 0,\,$ i.e., $t^2(t-1)(t+11)\le 0,\,$ which is true since $t\in [0,1].\,$ Equality is attained at $(1,1,1),\,$ $(3,0,0)\,$ and permutations. Returning to the original notations, equality is attained at $\displaystyle \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right),\,$ $(1,0,0)\,$ and permutations.

$\mathbf{\text{Case }\lambda\gt 26.}$

Using the above, suffice it to prove that $a^4+b^4+c^4+\lambda abc-\displaystyle \frac{\lambda+1}{27}\le a^4+b^4+c^4+26abc-1,\,$ for $\lambda \ge 26.\,$ This is equivalent to $\displaystyle (\lambda -26)abc\le\frac{\lambda -26}{27},\,$ which is true because $\lambda-26\ge 0\,$ and, by the AM-GM inequality, $\displaystyle abc\le\frac{1}{27}.$

For $\lambda\gt 26,\,$ equality is attained at $\displaystyle \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right);\,$ for $\lambda =26,\,$ at $\displaystyle \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right),\,$ $(1,0,0)\,$ and permutations, as above.

### Solution 2

Intuition (no proof): lets $lhs=lhs1+lhs2:$

1. $lhs1=a^4+b^4+c^4\in\displaystyle \left[\frac{1}{27},\right].\,$ (convexity argument, minimum for $a=b=c,\,$ maximum fir either variable $=1\,$ and the others $0)$.

2. $lhs2=\lambda abc\in\displaystyle \left[0,\frac{1}{27}\right],\,$ (since by the AM-GM inequality $\displaystyle (abc)^{\frac{1}{3}}\le\frac{1}{3}(a+b+c)=\frac{1}{3}.$

When $lhs2\,$ is at its maximum, $\displaystyle lhs=\frac{\lambda+1}{27};\,$ or $lhs1\,$ is at its maximum, $lhs=1.\,$ (We cannot prove for intermediate values without calculus.)

Proof by calculus: let $f=a^4+b^4+(a+b-1)^4-\lambda ab(a+b-1).\,$

\displaystyle \begin{align} \frac{\partial f}{\partial a}=4a^3-ab\lambda+b\lambda (1-a-b)-4(1-a-b)^3=0,\\ \frac{\partial f}{\partial b}=a\lambda (1-a-b)-ab\lambda-4(1-a-b)^3+4b^3=0 \end{align}

give for only solution satisfying the constraints $\displaystyle a=b=c=\frac{1}{3},\,$ hence $\displaystyle lhs=\frac{1}{27}+\frac{\lambda}{27}\,$ which satisfies the inequality.

### Acknowledgment

The problem and his solution have been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. Leo credits the part with $\lambda =26\,$ to Hung Nguyen Viet, the rest $(\lambda\gt 26)\,$ to Marin Chirciu. Solution 2 is by N. N. Taleb.