1) by taking (a,b,c,d)=(t,t,t,1/t^3) we get that: \[ 3t^x+t^{-3x}>=3/t+t^3 \] this inequality is true only when x>=3 (use: t->+infinity) 2) for x=3: first approach: apply AM-GM to \[ a^3+b^3+c^3>=3abc = 3/d \] ....(an alternative solution is by using the rearrangement inequality directly!)

What about $x=-1$?

methinks the problem is for x>=0. otherwise it is obvious!

bel.jad5 wrote: me thinks the problem is for x>=0. otherwise it is obvious! I solved the given problem: Olemissmath wrote: Let $a,b,c,d$ be positive real numbers such that $abcd=1$.Find with proof that $x=3 $ is the minimal value for which the following inequality holds : $a^x+b^x+c^x+d^x\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}$ By the way, we can say that all solved problems are obvious. We can say...

it is more rewarding to think about hard and unsolved problems... So assum that x>=0, can you come up with another approach?

Express very well : Let a, b, c and d be positive real numbers such that $abcd=1.$ Then prove that$ a^{x} + b^{x} + c^{x} + d^{x} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \Rightarrow x\geq3.$

sqing wrote: Express very well : Let a, b, c and d be positive real numbers such that $abcd=1.$ Then prove that$ a^{x} + b^{x} + c^{x} + d^{x} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \Rightarrow x\geq3.$ But it's wrong! See one of my previous posts.

arqady wrote: sqing wrote: Express very well : Let a, b, c and d be positive real numbers such that $abcd=1.$ Then prove that$ a^{x} + b^{x} + c^{x} + d^{x} \geq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \Rightarrow x\geq3.$ But it's wrong! See one of my previous posts. Pls can you send me a correct solution for this inequality problem ?

but i think it is not so easy as it looked like.i haven't ever found out why 3 is the minimal value and how to proof 3 is OK seriously.

who has the official solution?

i'm always looking forward.

The solution is here since 2013:( valid only for $x\geq0$) bel.jad5 wrote: 1) by taking $(a,b,c,d)=(t,t,t,\frac{1}{t^3})$ we get that: \[ 3t^x+t^{-3x}>=3/t+t^3 \]this inequality is true only when $x\geq3$ (use: t->+infinity) 2) for x=3: first approach: apply AM-GM to \[ a^3+b^3+c^3>=3abc = 3/d \]....(an alternative solution is by using the rearrangement inequality directly!)

Excuse me.Could you please tell me why when a^3+b^3+c^3>=3/d,it's obvious that the original inequality is true.And I'm interested in using the rearrangement inequality directly,but I cnnot work it out.

bobaboby1 wrote: Excuse me.Could you please tell me why when a^3+b^3+c^3>=3/d,it's obvious that the original inequality is true.And I'm interested in using the rearrangement inequality directly,but I cnnot work it out. In step 1, I proved that: $x \geq 3$. Now I will prove that $x=3$ is a solution. so by AM-GM: \[a^3+b^3+c^3 \geq 3\sqrt[3]{a^3b^3c^3}=3abc=\frac{3}{d}\]\[a^3+c^3+d^3 \geq 3\sqrt[3]{a^3c^3d^3}=3acd=\frac{3}{b}\]\[a^3+b^3+d^3 \geq 3\sqrt[3]{a^3b^3d^3}=3abd=\frac{3}{c}\]\[b^3+c^3+d^3 \geq 3\sqrt[3]{b^3c^3d^3}=3bcd=\frac{3}{a}\] Now if we sum all these inequalities, we get: \[3(a^3+b^3+c^3+d^3) \geq 3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})\]Then finally: \[a^3+b^3+c^3+d^3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\]

bel.jad5 wrote: 1) by taking (a,b,c,d)=(t,t,t,1/t^3) we get that: \[ 3t^x+t^{-3x}>=3/t+t^3 \]this inequality is true only when x>=3 (use: t->+infinity) 2) for x=3: first approach: apply AM-GM to \[ a^3+b^3+c^3>=3abc = 3/d \]....(an alternative solution is by using the rearrangement inequality directly!) please guys helpp please i didn't really understand how he proceed to prove the desired inequality why did he look for an example and the prove that 3 works ? why the example is necessary ?

琴声不等式可以解决，因为只需要二阶导大于0而在小于三很容易证明不对