# Show: $(x+y)^4 \leq 8(x^4 + y^4)$

I wish to show the following statement:

$\forall x,y \in \mathbb{R}$

$$(x+y)^4 \leq 8(x^4 + y^4)$$

What is the scope for generalisaion?

Edit:

Apparently the above inequality can be shown using the Cauchy-Schwarz inequality. Could someone please elaborate, stating the vectors you are using in the Cauchy-Schwarz inequality:

$\ \ \forall \ \ v,w \in V,$ an inner product space,

$$|\langle v,w\rangle|^2 \leq \langle v,v \rangle \cdot \langle w,w \rangle$$

where $\langle v,w\rangle$ is an inner product.

• homogeneous. Enough to prove for fixed $y=1$ and then $y=0.$ Nov 19, 2014 at 17:55

Apply Cauchy-Schwarz inequality twice: $x^4 + y^4 \geq \dfrac{1}{2}\left(x^2+y^2\right)^2 \geq \dfrac{1}{2}\left(\dfrac{1}{2}\left(x+y\right)^2\right)^2 = \dfrac{1}{8}\left(x+y\right)^4$.

• Please could you elaborate on you answer stating your choice of vectors in the Cauchy-Schwarz inequality: $|\langle v,w\rangle| \leq \langle v,v \rangle \cdot \langle w,w \rangle$ Nov 20, 2014 at 14:09
• @DanielKelsall choose $u=(1,1)$ and $v=(x,y)$ to get $(x+y)^2\le 2(x^2+y^2)$. Nov 21, 2014 at 11:28

A more general result is ($x,y\geq 0$, $p\geq 1$) $$(x+y)^p \leq 2^{p-1} (x^p+y^p),$$ which is direct consequence of convexity of $t\mapsto t^p$.

Regarding your edit and the question in the comment under OC-Sansoo's answer: (If I understand your issue right, you want reasoning for the choice of vectors?)

Start with the RHS of the inequality we want to show.

$$8\left(x^4+y^4\right) = \left(x^4+y^4\right)\left(2^2+2^2\right)$$ On the RHS we now have have the vectors $\vec{v}=(x^2,y^2)$ and $\vec{w}=(2,2)$.

Now we apply the CS inequality the first time: $$\left(2x^2+2y^2\right)^2 \leq \left(x^4+y^4\right)\left(2^2+2^2\right)$$ We do the same procedure again with the LHS term in the bracket (the inner product of the vectors $\vec{v}$ and $\vec{w}$): $$2x^2+2y^2= (1+1)(x^2+y^2)$$ Here we have the vectors $\vec{v}=(1,1)$ and $\vec{w}=(x,y)$.

Applying CS again: $$\left(x+y\right)^2 \leq (1+1)(x^2+y^2)$$

Now we are done, since $(x+y)^4\leq\left(2x^2+2y^2\right)^2$.

On a side note: In your edit the CS inequality should be: $$|\langle v,w\rangle|^2 \leq \langle v,v \rangle \cdot \langle w,w \rangle$$

• Thanks for the great answer. I'd never of thought of using Cauchy-Schwarz in this context before. Nov 21, 2014 at 21:45

If you instead consider $$\left( \frac{x}{2} + \frac{y}{2} \right)^4$$ we know that the function $(\cdot)^4$ is convex. This leads to: $$\left( \frac{x}{2} + \frac{y}{2} \right)^4 \le \frac12 x^4 + \frac12 y^4$$

Multiply both sides by $16$ and we have: $$(x+y)^4 \le 8x^4 + 8y^4.$$

This process works as long as $(\cdot)^p$ is convex, which holds precisely when $p \ge 1$.

You can show that $(x+y)^p \le x^p + y^p$ when $p < 1$ by other means.

we have $8(x^4+y^4)-(x+y)^4=7x^4-4x^3y-6x^2y^2-4xy^3+7y^4=(7x^2+10xy+7y^2)(x-y)^2\geq 0$ this is true.