How Can We Prove \(a+b^2+c^3+d^4 \le a^2+b^3+c^4+d^5\) Given \(a+b+c+d=4\)?

[anemone]

Let $a,\,b,\,c$ and $d$ be non-negative and suppose that $a+b+c+d=4$. Show that

$a+b^2+c^3+d^4\le a^2+b^3+c^4+d^5$

 

[Albert1]

Let $a,\,b,\,c$ and $d$ be non-negative and suppose that $a+b+c+d=4$. Show that$a+b^2+c^3+d^4\le a^2+b^3+c^4+d^5$

Spoiler

if $a=b=c=d=1 $ ,then $a+b^2+c^3+d^4= a^2+b^3+c^4+d^5=a+b+c+d=4$
now we only have to prove : $a^2+b^3+c^4+d^5- a -b^2 -c^3 - d^4\geq 0$
or :$a(a-1)+b^2(b-1)+c^3(c-1)+d^4(d-1)\geq 0 ----(1)$
if $a<1, b<1,c<1$ then $d>1$
we have: $(1)>(a-1)+(b-1)+(c-1)+d^4(d-1)=1-d+d^4(d-1)=(d-1)(d^4-1)>0$
for other situations the proof is similar, and the proof is done

 

[anemone]

Thanks for participating, Albert! Yes, your proof is correct! :)

Suggested solution of other:

Spoiler

We want to show that

$a^2+b^3+c^4+d^5-(a+b^2+c^3+d^4)\ge 0$

The LHS of the inequality is equivalent to

$a(a-1)+b^2(b-1)+c^3(c-1)+d^4(d-1)$

And we want to show it is non-negative.

But observe that

$d^4(d-1)-(d-1)=(d^4-1)(d-1)=(d-1)^2(d^3+d^2+d+1)\ge 0$

Thus,

$d^4(d-1)\ge (d-1)$

Similarly, we have $c^3(c-1)\ge (c-1)$, b^2(b-1)\ge (b-1)$ and a(a-1)\ge (a-1)$.

We conclude therefore that

$a(a-1)+b^2(b-1)+c^3(c-1)+d^4(d-1)\ge (a-1)+ (b-1)+ (c-1)+ (d-1)=0$