2021 Philippine MO

any ideas?

From the hypothesis $$43-a^2 \ge \frac{(13-a)^2}{3}$$ we get $5/2 \le a \le 4$ and so $$\frac{5}{2} \le d \le c \le b \le a \le 4.$$ We have $$(a+b)^2+(c+d)^2=43+2(ab+cd) \Leftrightarrow (13-c-d)^2+(c+d)^2=43+2(ab+cd)$$ So we must prove $$(13-c-d)^2+(c+d)^2 \ge 43+2(2cd+3).$$ This is equivalent to $$(c-d)^2 \ge (c+d-6)(20-c-d).$$ So it suffices to prove $$c+d \le 6 \Leftrightarrow a+b \ge 7.$$ The inequality $(b-c)(b-d) \ge 0$ gives $$3b^2+(c+d)^2+2cd \ge (b+c+d)^2.$$ Now by replacing $$c+d=13-a-b, \ cd = a^2+ab+b^2-13(a+b)+63$$ we will get $$3b^2-2(13-a)b + a^2-13a+63 \ge 0.$$ Solve this as a quadratic polynomial of $b$ with the notation $$b \ge \frac{13-a}{3} \ \ (a+3b \ge a+b+c+d =13)$$ You will obtain $$a \ge b \ge \frac{13 - a + \sqrt{(4-a)(2a-5)}}{3}.$$ This leads to $7/2 \le a \le 4$ and now the inequality $$a + \frac{13 - a + \sqrt{(4-a)(2a-5)}}{3} \ge 7$$ is satisfied. Equality holds for $(4,3,3,3).$

Doxuantrong wrote: From the hypothesis we can easily obtain (by AM - GM) $$\frac{5}{2} \le d \le c \le b \le a \le 4.$$ We have $$(a+b)^2+(c+d)^2=43+2(ab+cd) \Leftrightarrow (13-c-d)^2+(c+d)^2=43+2(ab+cd)$$ So we must prove $$(13-c-d)^2+(c+d)^2 \ge 43+2(2cd+3).$$ This is equivalent to $$(c-d)^2 \ge (c+d-6)(20-c-d).$$ So it suffices to prove $$c+d \le 6 \Leftrightarrow a+b \ge 7.$$ The inequality $(b-c)(b-d) \ge 0$ gives $$3b^2+(c+d)^2+2cd \ge (b+c+d)^2.$$ Now by replacing $$c+d=13-a-b, \ cd = a^2+ab+b^2-13(a+b)+63$$ we will get $$3b^2-2(13-a)b + a^2-13a+63 \ge 0.$$ Solve this as a quadratic polynomial of $b$ with the notation $$b \ge \frac{13-a}{3} \ \ (a+3b \ge a+b+c+d =13)$$ You will obtain $$a \ge b \ge \frac{13 - a + \sqrt{(4-a)(2a-5)}}{3}.$$ This leads to $7/2 \le a \le 4$ and now the inequality $$a + \frac{13 - a + \sqrt{(4-a)(2a-5)}}{3} \ge 7$$ is satisfied. why $\frac{5}{2} \le d \le c \le b \le a \le 4.$ ?

Let $a \ge b \ge c \ge d$ be real numbers such that $$a+b+c+d=13, \ a^2+b^2+c^2+d^2=43.$$ Prove that for all $k \in \mathbb{R}$ $$\min \left \{4k+3; \frac{7k+7}{2} \right \} \le ka+b \le \frac{13k+13 + \sqrt{9k^2-6k+9}}{4}.$$ When does equality hold ?

Nice inequality and solution.

go_placidly wrote: Let $a, b, c,$ and $d$ be real numbers such that $a \geq b \geq c \geq d$ and $$a+b+c+d = 13$$ $$a^2+b^2+c^2+d^2=43.$$ Show that $ab \geq 3 + cd$. There is a simplier proof. The key is, $$\begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}$$ which gives us $ab+cd \ge 21,$ or $$(a+b)^2+(c+d)^2 \ge a^2+b^2+c^2+d^2+42,$$ $$(a+b)^2+(13-a-b)^2 \ge 85,$$ $$(a+b-6)(a+b-7) \ge 0,$$ $$a+b \ge 7.$$ Hence, $ab-cd \ge ab - \frac{c^2+d^2}{2} =ab + \frac{a^2+b^2-43}{2} =\frac{(a+b)^2-43}{2} \ge 3.$

Let $\frac{39}{10}\ge a \geq b \geq c \geq d\geq 0, a+b+c+d = 13,a^2+b^2+c^2+d^2=43.$ Show that $$ab-cd \ge \frac 7{2}.$$

sqing wrote:

Dear sir, I really respect you. But if you think a problem/solution is beautiful then kindly upvote the problem/solution rather then writing this.

IMO Shortlist 2005 problem A3: Four real numbers $p$, $q$, $r$, $s$ satisfy $p+q+r+s = 9$ and $p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $\left(a,b,c,d\right)$ of $\left(p,q,r,s\right)$ such that $ab-cd \geq 2.$

First we show it suffices to prove that $a+b\geq7$. $$2ab+43=(a+b)^2+c^2+d^2\geq 49+c^2+d^2\geq 49+2cd.$$ Therefore, $$ab\geq 3+cd.$$ We now prove that $a+b\geq 7$. Let $a+b=7-t$ and $a^2+b^2=s$. We show that $t\leq0$. (Clearly $t\leq1/2$) We have $c+d=6+t$ and $c^2+d^2=43-s$. We solve for $a,b,c,d$ in terms of $s,t$ and apply the condition that $b\geq c$ to get $$(7-t)-\sqrt{2s-(7-t)^2}\geq (6+t)+\sqrt{86-2s-(6+t)^2}$$ rearrange and square both sides (can also see here again that $t\leq 1/2$) $$(1-2t)^2\geq 1+2t-2t^2+2\sqrt{(2s-(7-t)^2)(86-2s-(6+t)^2)},$$ $$3t(t-1)\geq \sqrt{(2s-(7-t)^2)(86-2s-(6+t)^2)}\geq0.$$ Therefore, as $t\leq1/2$ and $t(t-1)\geq0$, we get that $t\leq 0$.

KaiRain wrote: go_placidly wrote: Let $a, b, c,$ and $d$ be real numbers such that $a \geq b \geq c \geq d$ and $$a+b+c+d = 13$$ $$a^2+b^2+c^2+d^2=43.$$ Show that $ab \geq 3 + cd$. There is a simplier proof. The key is, $$\begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}$$ which gives us $ab+cd \ge 21,$ or $$(a+b)^2+(c+d)^2 \ge a^2+b^2+c^2+d^2+42,$$ $$(a+b)^2+(13-a-b)^2 \ge 85,$$ $$(a+b-6)(a+b-7) \ge 0,$$ $$a+b \ge 7.$$ Hence, $ab-cd \ge ab - \frac{c^2+d^2}{2} =ab + \frac{a^2+b^2-43}{2} =\frac{(a+b)^2-43}{2} \ge 3.$ How do you think of $$\begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}$$ at first?

qinghong wrote: How do you think of $$\begin{align*} a^2+b^2+c^2+d^2+6\left(ab+cd \right) & = \left(a+b+c+d\right)^2+2\left(a-c\right)\left(b-d\right)+2\left(a-d\right)\left(b-c\right) \\ & \ge \left(a+b+c+d\right)^2,\end{align*}$$ at first? The motivation is finding a relation betwen $a+b+c+d, a^2+b^2+c^2+d^2, ab$ and $cd$. We know: $$\begin{align*} \frac{(a+b+c+d)^2-(a^2+b^2+c^2+d^2)}{2} = \left(ab+cd \right) + \left(ac+bd \right) + \left(ad +bc \right) ,\end{align*}$$ which suggests us to prove: $$\begin{align*} 3 \left(ab+cd \right) \ge \left(ab+cd \right) + \left(ac+bd \right) + \left(ad +bc \right). \end{align*}$$