Irish Mathematical Olympiad

We have $$(a)(6,0,0,0)>(2,2,1,1)$$ $$(b)(2,2,2,0)>(2,2,1,1)$$ $$(c)(4,2,0,0)>(2,2,1,1)$$ By Muirhead and the above majorizations we have the following $$1.6(a^6+b^6+c^6+d^6)\ge 4(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ $$2.6(a^2{b}^2{c}^2+a^2{b}^2{d}^2+a^2{c}^2{d}^2+b^2{c}^2{d}^2)\ge 4(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ $$3(a^4{d}^2+a^2{b}^4+a^2{d}^4+b^4{d}^2+b^2{d}^4+a^4{c}^2+a^2{c}^4+b^4{c}^2+b^2{c}^4+c^4{d}^2+c^2{d}^4+b^2{a}^4)\ge 4(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ And so we have $$1^{\prime }(a^6+b^6+c^6+d^6)\ge \frac{4}{6}(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ $$2^{\prime }(a^2{b}^2{c}^2+a^2{b}^2{d}^2+a^2{c}^2{d}^2+b^2{c}^2{d}^2)\ge \frac{4}{6}(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ $$3^{\prime }(a^4{d}^2+a^2{b}^4+a^2{d}^4+b^4{d}^2+b^2{d}^4+a^4{c}^2+a^2{c}^4+b^4{c}^2+b^2{c}^4+c^4{d}^2+c^2{d}^4+b^2{a}^4)\ge 2(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2)$$ Taking the linear combination $(1^{\prime })+6(2^{\prime })+3(3^{\prime })$ we have it that l.h.s. of the resulting inequality becomes $(a^2+b^2+c^2+d^2{)}^3$ and thus $$(a^2+b^2+c^2+d^2{)}^3\ge \frac{32}{3}(a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2).$$ As $a^2+b^2+c^2+d^2=1$ the result follows.

Solution 2:

Writing the expression $a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2$ as $abcd(ab+bc+cd+ac+ad+bd)$ and given the condition $a^2+b^2+c^2+d^2=1$ we note that $abcd$ can not exceed $1/16$. Furthermore, using non-negativity of square of $(a-b),(b-c),(c-d),(d-a),(c-a),(d-b)$, and the condition $a^2+b^2+c^2+d^2=1$ yields that $ab+bc+cd+ac+ad+bd$ can not exceed $3/2$, and result follows.

Solution 3:

As above, we have $$a^2{b}^{2}cd+a{b}^2{c}^{2}d+ab{c}^2{d}^2+a^{2}b{c}^{2}d+a^{2}bc{d}^2+a{b}^{2}c{d}^2=abcd(ab+bc+cd+ac+ad+bd)$$ and $abcd\le \frac{1}{16}$ (by AM-GM and $a^2+b^2+c^2+d^2=1$). Now $$ab+bc+cd+ac+ad+bd\le \frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{c^2+d^2}{2}+\frac{a^2+c^2}{2}+\frac{a^2+d^2}{2}+\frac{b^2+d^2}{2}$$ (apply AM-GM to each term of the left hand side). Now the right hand side of the last inequality is equal to $\frac{3}{2}(a^2+b^2+c^2+d^2)$ which, by assumption is $\frac{3}{2}$. The required inequality follows immediately from these observations.