$2 (a^3 + b^3 + c^3) < (a + b + c) (a^2 + b^2 + c^2) $ $\iff a^3 + b^3 + c^3 < a^2( b + c)+b^2( c + a)+c^2( a + b )$ $\iff 0 < a^2( b + c-a)+b^2( c + a-b)+c^2( a + b-c )$ (use triangle inequalities) $(a + b + c) (a^2 + b^2 + c^2) \le 3 (a^3 + b^3 + c^3)$ $\iff a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \le 2(a^3 + b^3 + c^3)$ (use AM-GM)

Let $a, b$ be positive real numbers. Prove that $$a^3 + b^3< (a + b ) (a^2 + b^2 ) \le 2 (a^3 + b^3 )$$

sqing wrote: Let $a, b$ be positive real numbers. Prove that $$a^3 + b^3< (a + b ) (a^2 + b^2 ) \le 2 (a^3 + b^3 )$$ Solution. Since $a,b>0$, one has that \begin{align*}&a^2 -ab+ b^2< a^2 + b^2 \le 2 (a^2 -ab+ b^2)\\ \Longrightarrow&(a+b)\left(a^2 -ab+ b^2\right)<(a+b)\left(a^2 + b^2\right) \le 2(a+b)(a^2 -ab+ b^2)\\ \Longrightarrow&a^3 + b^3< (a + b ) (a^2 + b^2 ) \le 2 (a^3 + b^3 ).\end{align*}We are done. $\blacksquare$

ytChen wrote: sqing wrote: Let $a, b$ be positive real numbers. Prove that $$a^3 + b^3< (a + b ) (a^2 + b^2 ) \le 2 (a^3 + b^3 )$$ Solution. Since $a,b>0$, one has that \begin{align*}&a^2 -ab+ b^2< a^2 + b^2 \le 2 (a^2 -ab+ b^2)\\ \Longrightarrow&(a+b)\left(a^2 -ab+ b^2\right)<(a+b)\left(a^2 + b^2\right) \le 2(a+b)(a^2 -ab+ b^2)\\ \Longrightarrow&a^3 + b^3< (a + b ) (a^2 + b^2 ) \le 2 (a^3 + b^3 ).\end{align*}We are done. $\blacksquare$ Thanks.

How can you use the AM-GM inequality to show that (a+b+c)(a^2 + b^2 + c^2) < 3(a^3 + b^3 + c^3)

#3

$\Leftrightarrow a^3+b^3<a^3+b^3+a^2b+ab^2\le2\left(a^3+b^3\right)$ The left inequality is obvious, and the right one is true iff $a^3+b^3\ge a^2b+ab^2$, obvious by Rearrangment. It's actually true for all reals.

#6

Assuming $a,b,c\in\mathbb R^+$ since you wanted AM-GM. $\Leftrightarrow ab^2+ac^2+ba^2+bc^2+ca^2+cb^2\le2a^3+2b^3+2c^3$, which factors as $\sum_{\text{cyc}}(a-b)^2(a+b)\ge0$, equality for the first addend when $a=b$ and $a=-b$, hence $a=b=c=0$ (absurd), so the inequality is strict.

Thank you

The RHS is just chebycheff inequality and LHS can be done by using triangle inequality

FadhbSolutionneur42 wrote: How can you use the AM-GM inequality to show that (a+b+c)(a^2 + b^2 + c^2) < 3(a^3 + b^3 + c^3) To do this purely using AM-GM, we can do something like this. Expand the inequality so that we have to show $$ ab^2 + ac^2 + ba^2 + bc^2 + ca^2 + cb^2 \leq 2(a^3 + b^3 + c^3). $$We can rewrite this as $$ (a^3 + b^3 - ab^2 - ba^2) + (b^3 + c^3 - cb^2 - bc^2) + (c^3 + a^3 - ca^2 - ac^2) \geq 0, $$so it suffices to show that $a^3 + b^3 - ab^2 - ba^2 \geq 0$ and then we can just sum this for different variables. To do this with AM-GM, we have a standard trick where you manually weight each of the variables, as shown below. $$ \frac{1}{3}a^3 + \frac{1}{3}b^3 + \frac{1}{3}b^3 \geq 3 \sqrt[3]{\frac{1}{3^3}a^3 b^3 b^3} = ab^2, $$and $$ \frac{1}{3}b^3 + \frac{1}{3}a^3 + \frac{1}{3}a^3 \geq 3 \sqrt[3]{\frac{1}{3^3}b^3 a^3 a^3} = ba^2, $$and adding these gives $a^3 + b^3 \geq ab^2 + ba^2$, which is what we needed. (We can ignore the equality case because $a, b, c$ are the sides of a triangle)