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

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The discussion centers on proving the inequality \(a+b^2+c^3+d^4 \le a^2+b^3+c^4+d^5\) under the constraint \(a+b+c+d=4\) with non-negative variables. Participants confirm the validity of a proof presented by one member, emphasizing the correctness of the approach taken. Various methods and strategies for proving the inequality are suggested, highlighting the importance of manipulating the terms based on the given condition. The conversation reflects a collaborative effort to explore mathematical reasoning and proofs. The thread concludes with a consensus on the proof's accuracy and the effectiveness of the discussed methods.
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Let $a,\,b,\,c$ and $d$ be non-negative and suppose that $a+b+c+d=4$. Show that

$a+b^2+c^3+d^4\le a^2+b^3+c^4+d^5$
 
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anemone said:
Let $a,\,b,\,c$ and $d$ be non-negative and suppose that $a+b+c+d=4$. Show that$a+b^2+c^3+d^4\le a^2+b^3+c^4+d^5$
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
 
Thanks for participating, Albert! Yes, your proof is correct! :)

Suggested solution of other:

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

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