一道高一简单数学题求解.求证a^4+b^4+c^4大于等于abc(a+b+c)

a^4+b^4+c^4
=1/2(a^4+b^4+b^4+c^4+c^4+a^4)
＝1/2[(a^4－2a^2×b^2+b^4+b^4-2b^2×c^2+c^4-2c^2×a^2+c^4+a^4)+2a^2×b^2+2b^2×c^2+2c^2×a^2]
=1/2[(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2]+a^2×b^2+b^2×c^2+c^2×a^2
因为[(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2]≥0
所以
原式＝a^4+b^4+c^4 ≥a^2×b^2+b^2×c^2+c^2×a^2
而同理
a^2×b^2+b^2×c^2+c^2×a^2
=1/2[a^2×b^2+b^2×c^2+b^2×c^2
＋c^2×a^2＋c^2×a^2 ＋a^2×b^2]
＝1/2[a^2×b^2－2acb^2+b^2×c^2+b^2×c^2-2abc^2
＋c^2×a^2＋c^2×a^2 -2bca^2＋a^2×b^2+2acb^2+2abc^2+2bca^2]
=1/2[(ab-bc)^2+(bc-ac)^2+(ac-ab)^2]+acb^2+abc^2+bca^2
=1/2[(ab-bc)^2+(bc-ac)^2+(ac-ab)^2]+abc(a+b+c)
因为[(ab-bc)^2+(bc-ac)^2+(ac-ab)^2]≥0
所以
a^2×b^2+b^2×c^2+c^2×a^2 ≥abc(a+b+c)
所以得证