Hint: We can say $abcd=1$, because this inequality homogenous.

FuadAnzurov2003 wrote: Hint: We can say $abcd=1$, because this inequality homogenous. Ok, then??

Using AM-GM: $$\sqrt{abcd} \le \frac{ab+cd}{2}$$; so we will prove: $$a^4+b^4+c^4+d^4 \ge 2(a-b)^2(ab+cd)+4abcd$$ we have: $$a^4+b^4+c^4+d^4=(a^2-b^2)^2+c^4+d^4+2a^2b^2$$ and $$(a^2-b^2)^2-2(a-b)^2ab=(a-b)^2(a^2+b^2) \ge \frac{1}{2}(a-b)^2(a+b)^2=\frac{1}{2}(a^2-b^2)^2$$ $$c^4+d^4 \ge 2c^2d^2$$ so we need $$\frac{1}{2}(a^2-b^2)^2+2a^2b^2+2c^2d^2 \ge 4abcd+2(a-b)^2cd$$ equivalent to $$\frac{(a^2+b^2)^2}{2}+2c^2d^2 \ge 2cd(a^2+b^2)$$ It is true by AM-GM. Done.

Mathuzb wrote: FuadAnzurov2003 wrote: Hint: We can say $abcd=1$, because this inequality homogenous. Ok, then?? We can write similarly $a^4+b^4+c^4+d^4 \geq 4abcd+(a-b)^2 \sqrt{abcd}+(b-c)^2 \sqrt{abcd}+(c-d)^2 \sqrt{abcd}+(d-a)^2 \sqrt{abcd}$. $abcd=1$ That's why $a^4+b^4+c^4+d^4 \geq 4+(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$ or $a^4+b^4+c^4+d^4+4 \geq 8+2(a^2+b^2+c^2+d^2-ab-bc-cd-da)$. With A.M-G.M $a^4+b^4+c^4+d^4+4 \geq 2(a^2+b^2+c^2+d^2)$ Then A.M-G.M $ab+bc+cd+da \geq 4$

FuadAnzurov2003 wrote: Mathuzb wrote: FuadAnzurov2003 wrote: Hint: We can say $abcd=1$, because this inequality homogenous. Ok, then?? We can write similarly $a^4+b^4+c^4+d^4 \geq 4abcd+(a-b)^2 \sqrt{abcd}+(b-c)^2 \sqrt{abcd}+(c-d)^2 \sqrt{abcd}+(d-a)^2 \sqrt{abcd}$. $abcd=1$ That's why $a^4+b^4+c^4+d^4 \geq 4+(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2$ or $a^4+b^4+c^4+d^4+4 \geq 8+2(a^2+b^2+c^2+d^2-ab-bc-cd-da)$. With A.M-G.M $a^4+b^4+c^4+d^4+4 \geq 2(a^2+b^2+c^2+d^2)$ Then A.M-G.M $ab+bc+cd+da \geq 4$ Why can write similarly?

Trunglindan1995 wrote: Using AM-GM: $$\sqrt{abcd} \le \frac{ab+cd}{2}$$; so we will prove: $$a^4+b^4+c^4+d^4 \ge 2(a-b)^2(ab+cd)+4abcd$$ we have: $$a^4+b^4+c^4+d^4=(a^2-b^2)^2+c^4+d^4+2a^2b^2$$ and $$(a^2-b^2)^2-2(a-b)^2ab=(a-b)^2(a^2+b^2) \ge \frac{1}{2}(a-b)^2(a+b)^2=\frac{1}{2}(a^2-b^2)^2$$ $$c^4+d^4 \ge 2c^2d^2$$ so we need $$\frac{1}{2}(a^2-b^2)^2+2a^2b^2+2c^2d^2 \ge 4abcd+2(a-b)^2cd$$ equivalent to $$\frac{(a^2+b^2)^2}{2}+2c^2d^2 \ge 2cd(a^2+b^2)$$ It is true by AM-GM. Done. good solutionTrunglindan1995! thank you

It was wrong.

Trunglindan1995 wrote: Why can write similarly? Because we don't have any condition about a,b,c,d min or max.

Mathuzb wrote: Let $a,b,c,d>0$ . Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge 4(a-b)^2 \sqrt{abcd}$$ The following inequality is a bit stronger: Let $a,b,c,d>0$ . Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$$

MariusStanean wrote: Mathuzb wrote: Let $a,b,c,d>0$ . Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge 4(a-b)^2 \sqrt{abcd}$$ The following inequality is a bit stronger: Let $a,b,c,d>0$ . Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$$ Anyone?

Bump on the second above post

MariusStanean wrote: The following inequality is a bit stronger: Let $a,b,c,d>0$ . Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$$ It suffices to prove $a^4+b^4+2c^2d^2 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$. $\bullet$ If $ab\ge cd$ then $L-R=(a-b)^4+(a-b)^2(3\sqrt{ab}+\sqrt{cd})(\sqrt{ab}-\sqrt{cd})+2(ab-cd)^2\ge 0$ $\bullet$ If $ab< cd$ then $L-R=\frac{1}{4}\left(2(a-b)^2+(3\sqrt{ab}+\sqrt{cd})(\sqrt{ab}-\sqrt{cd})\right)^2+\frac{1}{4}(\sqrt{ab}-\sqrt{cd})^2\left(7cd-ab+10\sqrt{abcd}\right)\ge 0$

Let $a, b, c, d > 0$. Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$$\[ a^4 + b^4 + c^4 + d^4 \geq 4abcd \left( 1 + \frac{(a-b)^2}{ab\sqrt{2}} + \frac{(c-d)^2}{cd\sqrt{2}} \right) + 2(a-b)^2(c-d)^2 \]Prove that for any real numbers $a,b,c,d,$ $$a^4+b^4+c^4+d^4-4abcd \ge 2|(a^2-b^2+c^2-d^2)(ab-cd)|.$$Prove that for all non-negative real numbers $a,b,c,d$, then $$a^4+b^4+c^4+d^4-4abcd \geq 2(a-b)(b-c)(c-d)(d-a).$$Let $a, b, c, d$ be positive real numbers. Prove that $$a^4+b^4+c^4+d^4-4abcd\ge \frac{1}{2}((a-b)^4+(b-c)^4+(c-d)^4+(d-a)^4)$$Let $a,b,c,d>0$. Show that: \[27(a^4+b^4+c^4+d^4-4abcd)\ge (a+b+c+d)^4-4^4abcd\]Let $a,b,c,d\geq 0.$ Prove that: \[a^4+c^4+b^4+d^4-4abcd\geq \frac{1}{128}(|a-b|+|b-c|+|c-d|+|d-a|)^4\]Let $a,b,c,d$ be positive numbers such that $a+b+c+d=2$.Prove that $$a^4+b^4+c^4+d^4-4abcd \ge 2(a-b)(c-d)$$here here here

sqing wrote: Let $a, b, c, d > 0$. Prove that $$a^4+b^4+c^4+d^4 - 4abcd\ge (a-b)^2 \left(\sqrt{ab}+\sqrt{cd}\right)^2$$\[ a^4 + b^4 + c^4 + d^4 \geq 4abcd \left( 1 + \frac{(a-b)^2}{ab\sqrt{2}} + \frac{(c-d)^2}{cd\sqrt{2}} \right) + 2(a-b)^2(c-d)^2 \]Prove that for any real numbers $a,b,c,d,$ $$a^4+b^4+c^4+d^4-4abcd \ge 2|(a^2-b^2+c^2-d^2)(ab-cd)|.$$Prove that for all non-negative real numbers $a,b,c,d$, then $$a^4+b^4+c^4+d^4-4abcd \geq 2(a-b)(b-c)(c-d)(d-a).$$Let $a, b, c, d$ be positive real numbers. Prove that $$a^4+b^4+c^4+d^4-4abcd\ge \frac{1}{2}((a-b)^4+(b-c)^4+(c-d)^4+(d-a)^4)$$Let $a,b,c,d>0$. Show that: \[27(a^4+b^4+c^4+d^4-4abcd)\ge (a+b+c+d)^4-4^4abcd\]Let $a,b,c,d\geq 0.$ Prove that: \[a^4+c^4+b^4+d^4-4abcd\geq \frac{1}{128}(|a-b|+|b-c|+|c-d|+|d-a|)^4\]Let $a,b,c,d$ be positive numbers such that $a+b+c+d=2$.Prove that $$a^4+b^4+c^4+d^4-4abcd \ge 2(a-b)(c-d)$$here here here beautiful collection!

please correct me if wrong * make a condition from the question

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evaluate the condition

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substitute the conditionClick to reveal hidden text

using the same technique we could prove the answers like the ones given by squig or/and teomihai and MariusStanean