Real numbers $a, b, c, d$ satisfy $$a^2+b^2+c^2+d^2=4.$$Find the greatest possible value of $$E(a,b,c,d)=a^4+b^4+c^4+d^4+4(a+b+c+d)^2 .$$
Problem
Source: Moldova TST 2022
Tags: inequalities
01.04.2022 16:56
Answer: $E(a,b,c,d)\le 68.$ Pf: Note that $x(-x+2)\le 1$ for all real $x$ and $s^2+t^2\ge 2st$ for all reals $s$ and $t$,we have \begin{align*} E(a,b,c,d)&=a^4+b^4+c^4+d^4+4(a+b+c+d)^2\\ &=(a^2+b^2+c^2+d^2)^2-2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)+4(a^2+b^2+c^2+d^2)+8(ab+ac+ad+bc+bd+cd) \\ &=32+2(ab(-ab+2)+ac(-ac+2)+ad(-ad+2)+bc(-bc+2)+bd(-bd+2)+cd(-cd+2))+4(ab+ac+ad+bc+bd+cd) \\ &\le 32+2\cdot 6+6(a^2+b^2+c^2+d^2)\\ &=32+2\cdot 6+6\cdot 4 \\ &=68. \ \blacksquare \end{align*}
01.04.2022 17:45
Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that $$4\leq a^4+b^4+c^4+d^4+4(a+b+c+d)^2\leq 68$$$$4\leq a^4+b^4+c^4+d^4+\frac{4}{3}(a+b+c+d)^2\leq \frac{76}{3}$$$$-60 \leq a^4+b^4+c^4+d^4+(a+b+c+d)^3\leq 68$$$$-12 \leq a^4+b^4+c^4+d^4+\frac{1}{4}(a+b+c+d)^3\leq 20$$$$-40\leq a^4+b^4+c^4+d^4+11(a+b+c+d)\leq 48$$$$-\frac{182}{5} \leq a^4+b^4+c^4+d^4+\frac{101}{10}(a+b+c+d)\leq \frac{222}{5}$$ augustin_p wrote: Real numbers $a, b, c, d$ satisfy $$a^2+b^2+c^2+c^2+d^2=4.$$Find the greatest possible value of $$E(a,b,c,d)=a^4+b^4+c^4+d^4+4(a+b+c+d)^2 .$$ https://artofproblemsolving.com/community/c6h1233291p6247673 https://artofproblemsolving.com/community/c6h1212723p6018660
18.04.2022 11:20
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=3.$ Prove that $$ -\frac{17}{\sqrt 2} \leq a^3+b^3+c^3+3(a+b+c)\leq\frac{17}{\sqrt 2}$$$$-\frac{23}{\sqrt 6} \leq a^3+b^3+c^3+2(a+b+c)\leq\frac{23}{\sqrt 6}$$$$-3<a^3+b^3+c^3+3(a+b+c)^2\leq30$$$$-4<a^3+b^3+c^3+\frac{2}{3}(a+b+c)^2\leq 9$$
18.04.2022 20:42
augustin_p wrote: Real numbers $a, b, c, d$ satisfy $$a^2+b^2+c^2+c^2+d^2=4.$$Find the greatest possible value of $$E(a,b,c,d)=a^4+b^4+c^4+d^4+4(a+b+c+d)^2 .$$ The problem is created by the great Romanian mathematician Ion Nedelcu and originally published by "G. Matematica"
19.04.2022 02:46
Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that$$-\frac{41}{2}\leq a^4+b^4+c^4+d^4+7(a+b+c+d+\frac{1}{2}abcd)\leq \frac{71}{2}$$Let $a,b,c$ be real numbers such that $a^2+b^2+c^2=3.$ Prove that $$ -\frac{59}{2\sqrt{17}}\leq a^3+b^3+c^3+a+b+c+\frac{1}{2}abc\leq\frac{59}{2\sqrt{17}}$$
19.04.2022 06:01
augustin_p wrote: Real numbers $a, b, c, d$ satisfy $$a^2+b^2+c^2+c^2+d^2=4.$$Find the greatest possible value of $$E(a,b,c,d)=a^4+b^4+c^4+d^4+4(a+b+c+d)^2 .$$ If $a,b,c,d$ are real numbers such that $$a^2+b^2+c^2+d^2=4,$$then $$3(a^4+b^4+c^4+d^4+12abcd)+80\geq8(a+b+c+d)^2.$$ (L. Giugiuc, 2022) https://artofproblemsolving.com/community/c6t243f6h2826863_regarding_m_164
21.04.2022 03:04
sqing wrote: Let $a,b,c,d$ be real numbers such that $a^2+b^2+c^2+d^2=4.$ Prove that $$4\leq a^4+b^4+c^4+d^4+4(a+b+c+d)^2\leq 68$$$$-60 \leq a^4+b^4+c^4+d^4+(a+b+c+d)^3\leq 68$$
Attachments:
