Prove that this inequality $5(a^2+b^2+c^2) \leq 6(a^3+b^3+c^3)+1$

Question

Let $a,b,c>0$ and $a+b+c=1$ Prove that $$5(a^2+b^2+c^2) \leq 6(a^3+b^3+c^3)+1$$

Answer 1

Since we are given $a+b+c =1$, normalize the entire inequality to

$$ 5 (a^2+b^2+c^2)(a+b+c) \leq 6(a^3+b^3+c^3) + (a+b+c)^3 $$

Now expand and simplify to get

$$ a^2b + b^2c + c^2a + a^2c + b^2a + c^2b \leq a^3 + b^3 + c^3 + 3abc $$

You should recognize this as Schurs's inequality.

Answer 2

Lagrange multipliers: consider the function $f(a,b,c)=6(a^3+b^3+c^3)+1-5(a^2+b^2+c^2)-\lambda(a+b+c-1).$ Note, that $\frac{d f}{da}=18a^2-10a-\lambda=0$ is a quadratic function with respect to $a$ and thus two numbers out $a,b,c$ must be equal. Letting $a=b=x$ and $c=1-2x$ the inequality can be rewritten in terms of one variable $x...$

The only case left is if one variable equals to $0.$ If $c=0,$ then our inequality becomes $6(b^3+c^3)+1\ge 5(b^2+c^2)$ and $b=1-c$ which again can be rewritten in terms of one variable. Alternatively, $b^3+c^3=(b+c)(b^2-bc+c^2)=b^2-bc+c^2,$ so we are left to show $b^2+c^2+1\ge 6bc,$ which is clearly true since $b^2+c^2\ge 2bc$ and $bc\le\frac{(b+c)^2}{4}=\frac{1}{4}.$

Answer 3

Let's try solving this using Lagrangian multiplier. Our goal is to prove that:

$$f(a,b,c) = 6(a^3 + b^3 + c^3) - 5(a^2 + b^2 + c^2) + 1$$

doesn't have negative values when this constrain holds:

$$g(a,b,c) = a + b + c - 1 = 0$$

$$F(a,b,c,\lambda) = 6(a^3 + b^3 + c^3) - 5(a^2 + b^2 + c^2) + 1 + \lambda(a + b + c - 1)$$

Now we take parital derivatives:

$$F_a = 18a^2 - 10a + \lambda = 0$$ $$F_b = 18b^2 - 10b + \lambda = 0$$ $$F_c = 18c^2 - 10c + \lambda = 0$$ $$F_\lambda = a + b + c - 1 = 0$$

From first two partial derivatives we have:

$$18a^2 - 10a = 18b^2 - 10b$$ $$18(a-b)(a+b) = 10(a-b)$$

Now we have 2 distinct cases:

$a-b \neq 0$

$$18(a+b) = 10$$ $$a+b = \frac{5}{9}$$

This implies that $c = \frac{4}{9}$, which later implies that $\lambda = -\frac{8}{9}$

Solving the two quadratic equations in first two partial derivatives gives two solution.

$$a_{1/2} = \frac{5 \pm \sqrt{41}}{18} b_{1/2} = \frac{5 \pm \sqrt{41}}{18} $$

Beacuse $a \neq b$ means that a or b is negative, which violate the condition.

$a-b = 0$

This implies $c= 1 - 2b$

Solving the second and the third partial derivative we have:

$$18b^2 - 10b = 18(1-2b)^2 - 10(1-2b)$$

Simplifying this lead to quadratic equation:

$$54b^2 - 42b + 8 = 0$$

This eqaution has two solution, which leads to 2 distinct cases:

$$b = a = \frac{4}{9}$$

This implies $c = \frac{1}{9}$

The function for this values have value of $f(\frac{4}{9},\frac{4}{9},\frac{1}{9}) = \frac{2}{81}$, which is a saddle point.

$$b = a = \frac{1}{3}$$

This implies $c = \frac{1}{3}$

The function for this values have value of $f(\frac{1}{3},\frac{1}{3},\frac{1}{3}) = 0$, which is the global manima.

Using Lagrange multiplier we prove that the inequality is true for $a+b+c = 1$ and the equality holds for $(a,b,c) = (\frac{1}{3},\frac{1}{3},\frac{1}{3})$

Q.E.D.

P.S. Withoud loss of generality we chose c to be treated differently that a and b, but beacuse the partial derivatives are homogeneous we don't have to do that to other variables separately