Q simplify
Note - use the easiest method _/\_
Answers
Answer:
I hope it is helpful for you ☺️
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Step-by-step explanation:
A big problem we get around (x,y,z)=(0.822,1.265,1.855).
The Buffalo Way helps:
Let x=min{x,y,z}, y=x+u,z=x+v and x=t
√
uv
.
Hence,
13
5
∏
cyc
(8x3+5y3)(
∑
cyc
x4
8x3+5y3
−
x+y+z
13
)=
=156(u2−uv+v2)x8+6(65u3+189u2v−176uv2+65v3)x7+
+2(377u4+1206u3v+585u2v2−1349uv3+377v4)x6+
+3(247u5+999u4v+1168u3v2−472u2v3−726uv4+247)x5+
+3(117u6+696u5v+1479u4v2+182u3v3−686u2v4−163uv5+117v6)x4+
+(65u7+768u6v+2808u5v2+2079u4v3−1286u3v4−585u2v5+181uv6+65v7)x3+
+3uv(40u6+296u5v+472u4v2−225u2v4+55uv5+25v6)x2+
+u2v2(120u5+376u4v+240u3v2−240u2v3−25uv4+75v5)x+
+5u3v3(8u4+8u3v−8uv3+5v4)≥
≥u5v5(156t8+531t7+2t6−632t5−152t4+867t3+834t2+299t+40)≥0
Done!
For example, we'll prove that
6(65u3+189u2v−176uv2+65v3)≥531
√
u3v3
,
which gives a coefficient 531 before t7 in the polynomial 156t8+531t7+2t6−632t5−152t4+867t3+834t2+299t+40.
Indeed, let u=k2v, where k>0.
Thus, we need to prove that:
130k6+378k4−177k3−352k2+130≥0
and by AM-GM we obtain:
130k6+378k4−177k3−352k2+130=
=130(k3+
10
13
k−1)2+
k
13
(2314k3+1079k2−5576k+2600)≥
≥
k
13
(8⋅
1157
4
k3+5⋅
1079
5
k2+21⋅
2600
21
−5576k)≥
≥
k2
13
(3434
√
(
1157
4
)8(
1079
5
)5(
2600
21
)21
−5576)>0.
We'll prove that
2(377u4+1206u3v+585u2v2−1349uv3+377v4)≥2u2v2,
for which it's enough to prove that:
377t4+1206t3+584t2−1349t+377≥0
or
t4+
1206
377
t3+
584
377
t2−
1349
377
t+1≥0
or
(t2+
603
377
t−
28
29
)2+
131015t2−69589t+9633
142129
≥0,
which is true because
695892−4⋅131015⋅9633<0.
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
Hi unnie I fine :-)
how are you
my studies are going well 〜(꒪꒳꒪)〜
just 2 exams are left (─.─||）
