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Prove: $28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4$ with $a, b, c \ge0$

I can do this by: $EAT^2$ (expand all of the thing)

  • $(x+y+z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y+4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{ x}^{2}{z}^{2}+4\,x{y}^{3}+4\,x{z}^{3}+4\,{y}^{3}z+6\,{y}^{2}{z}^{2}+4 \,y{z}^{3}+12\,x{y}^{2}z+12\,xy{z}^{2}+12\,{x}^{2}yz$

  • $(x+y-z)^4={x}^{4}+{y}^{4}+{z}^{4}+4\,{x}^{3}y-4\,{x}^{3}z+6\,{x}^{2}{y}^{2}+6\,{ x}^{2}{z}^{2}+4\,x{y}^{3}-4\,x{z}^{3}-4\,{y}^{3}z+6\,{y}^{2}{z}^{2}-4 \,y{z}^{3}-12\,x{y}^{2}z+12\,xy{z}^{2}-12\,{x}^{2}yz$

...

$$28(a^4+b^4+c^4)\ge (a+b+c)^4+(a+b-c)^4+(b+c-a)^4+(a+c-b)^4\\ \iff a^4 + b^4 + c^4 \ge a^2b^2+c^2a^2+b^2c^2 \text{(clearly hold by AM-GM)}$$

but any other ways that smarter ?

Xeing
  • 2,967

3 Answers3

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A nice way of tackling the calculations might be as follows:$$~$$ Let $x=b+c-a,y=c+a-b,z=a+b-c.$ Then the original inequality is just equivalent with $$\frac74\Bigl((x+y)^4+(y+z)^4+(z+x)^4\Bigr)\geq x^4+y^4+z^4+(x+y+z)^4.$$ Now we can use the identity $$\sum_{cyc}(x+y)^4=x^4+y^4+z^4+(x+y+z)^4-12xyz(x+y+z),$$ So that it suffices to check that $$\frac37\Bigl(x^4+y^4+z^4+(x+y+z)^4\Bigr)\geq 12xyz(x+y+z),$$ Which obviously follows from the AM-GM inequality: $(x+y+z)^4\geq \Bigl(3(xy+yz+zx)\Bigr)^2\geq 27xyz(x+y+z)$ and $x^4+y^4+z^4\geq xyz(x+y+z).$ Equality holds in the original inequality iff $x=y=z\iff a=b=c.$ $\Box$

Potla
  • 521
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This inequality is true for all reals $a$, $b$ and $c$.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.

Hence, our inequality is a linear inequality of $w^3$, which says that it's enough to prove our inequality for an extremal value of $w^3$, which happens for equality case of two variables.

Since our inequality is homogeneous and fourth degree, we can assume $b=c=1$,

which gives $(a-1)^2(a+1)^2\geq0$.

Done!

Michael Rozenberg
  • 194,933
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For non-negative variables also TL helps:

Let $a+b+c=3$ (we can assume it because our inequality is homogeneous).

Hence, we need to prove that $$\sum_{cyc}(28a^4-(3-2a)^4-27)\geq0$$ or $$\sum_{cyc}(a-1)(a^3+9a^2-9a+9)\geq0$$ or $$\sum_{cyc}\left((a-1)(a^3+9a^2-9a+9)-10(a-1)\right)\geq0$$ or $$\sum_{cyc}(a-1)^2(a^2+10a+1)\geq0.$$ Done!

Michael Rozenberg
  • 194,933