If $a+b+c=3,$ find max $T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}$

Question

Let $a,b,c\ge 0: a+b+c=3.$ Find the maximal value $$T=\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}.$$ By $a=b=c=1,$ I get $M=3$ and I tried to prove it is maximum.

It means to prove$$\sqrt{\frac{6a+7bc}{6a+7}}+\sqrt{\frac{6b+7ca}{6b+7}}+\sqrt{\frac{6c+7ab}{6c+7}}\le 3.$$ Now, I used C-S to eliminate the radical and it is enough to prove $$\frac{6a+7bc}{6a+7}+\frac{6b+7ca}{6b+7}+\frac{6c+7ab}{6c+7}\le 3.$$ But it is not true. I think there is some better approachs but I failured to find it.

All idea is welcome. Thanks.

Answer 1

By C-S $$\sum_{cyc}\sqrt{\tfrac{6a+7bc}{6a+7}}\leq\sqrt{\sum_{cyc}\tfrac{6a+7bc}{2a+21}\sum_{cyc}\tfrac{2a+21}{6a+7}}=\sqrt{3\sum_{cyc}\tfrac{2a^2+2ab+2ac+7bc}{9a+7b+7c}\sum_{cyc}\tfrac{9a+7b+7c}{25a+7b+7c}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{2a^2+2ab+2ac+7bc}{9a+7b+7c}\sum_{cyc}\frac{9a+7b+7c}{25a+7b+7c}\leq a+b+c$$ or $$\sum_{sym}(18350.5a^7+160622a^6b+89992a^5b^2-287315a^4b^3)+$$ $$+\sum_{sym}(279968.5a^5bc+179966a^4b^2c-228781a^3b^3c-212803a^3b^2c^2)\geq0.$$ Now, by Schur $$\sum_{sym}(a^7-2a^6b+a^5bc)\geq0,$$ which by Muirhead gives $$\sum_{sym}(18350.5a^7+160622a^6b+89992a^5b^2-287315a^4b^3)\geq$$ $$\geq\sum_{sym}(197323a^6b+89992a^5b^2-287315a^4b^3-18350.5a^5bc)\geq$$ $$\geq-\sum_{sym}18350.5a^5bc$$ and it's enough to prove that: $$\sum_{sym}(261618a^5bc+179966a^4b^2c-228781a^3b^3c-212803a^3b^2c^2)\geq0,$$ which is true by Muirhead.

Answer 2

Supplement to Michael Rozenberg's very nice answer.

It suffices to prove that $$\sum_{\mathrm{cyc}}\frac{6a+7bc}{2a+21}\sum_{\mathrm{cyc}}\frac{2a+21}{6a+7} \le 9. \tag{1}$$

We use the pqr method.

Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.

We have $$\sum_{\mathrm{cyc}}\frac{6a+7bc}{2a+21} = \frac{28q^2 + 4473q - 978r + 7938}{11907 + 84q + 8r}, $$ and $$\sum_{\mathrm{cyc}}\frac{2a+21}{6a+7} = \frac{8673 + 924q + 216r}{1225 + 252q + 216r}.$$

Thus, (1) is written as $$\frac{28q^2 + 4473q - 978r + 7938}{11907 + 84q + 8r} \cdot \frac{8673 + 924q + 216r}{1225 + 252q + 216r} \le 9$$ or $$\frac{28q^2 + 4473q - 978r + 7938}{11907 + 84q + 8r} \cdot \left(1 + \frac{7448 + 672q}{1225 + 252q + 216r}\right) \le 9. \tag{2}$$

If $q \le 9/4$, it suffices to prove that $$\frac{28q^2 + 4473q - 978\cdot 0 + 7938}{11907 + 84q + 8\cdot 0} \cdot \left(1 + \frac{7448 + 672q}{1225 + 252q + 216\cdot 0}\right) \le 9$$ or $$\frac{(4q-9)(44q^2 + 7217q + 47187)}{(567+4q)(175+36q)} \le 0$$ which is true.

If $9/4 < q \le 3$, using degree three Schur $r \ge \frac{4pq - p^3}{9} = \frac{4q - 9}{3} \ge 0$, it suffices to prove that $$\frac{28q^2 + 4473q - 978\cdot \frac{4q - 9}{3} + 7938}{11907 + 84q + 8\cdot \frac{4q - 9}{3}} \cdot \left(1 + \frac{7448 + 672q}{1225 + 252q + 216\cdot \frac{4q - 9}{3}}\right) \le 9$$ or $$\frac{63(q-3)(404q+45043)(4q-9)}{(35649+284q)(577+540q)}\le 0$$ which is true.

We are done.

Answer 3

More about the Cauchy-Bunyakovsky-Schwarz trick.

We want to find $p_1, p_2, p_3, p_4$ such that $$\sum_{\mathrm{cyc}}\frac{6a + 7bc}{p_1a + p_2 b + p_3 c + p_4} \sum_{\mathrm{cyc}}\frac{p_1a + p_2 b + p_3 c + p_4}{6a + 7} \le 9.$$ (Note: By symmetry, we can let $p_2 = p_3$.)

By the equality cases, it boils down to finding $p_1, p_2$ such that $$\sum_{\mathrm{cyc}}\frac{6a + 7bc}{p_1a + p_2 b + p_2 c + \frac{21}{2}p_1 - \frac{27}{2}p_2} \sum_{\mathrm{cyc}}\frac{p_1a + p_2 b + p_2 c + \frac{21}{2} - \frac{27}{2}p_2}{6a + 7} \le 9.$$

Using $b + c = 3- a$, we have $$p_1a + p_2 b + p_2 c + \frac{21}{2}p_1 - \frac{27}{2}p_2 = \frac12(p_1 - p_2)(2a + 21).$$

So we choose $2a + 21$.

Sometimes, we need a quadratic function $p_1 a^2 + p_2 b^2 + p_3 c^2 + p_4 ab + p_5bc + p_6ca$.

Problem 1 (by Ji Chen). Let $a, b, c > 0$. Prove that $$\sqrt{\frac{578a^2+1143bc}{527(a+b+c)^2}} + \sqrt{\frac{578b^2+1143ca}{527(a+b+c)^2}} + \sqrt{\frac{578c^2+1143ab}{527(a+b+c)^2}} \le \frac{253}{140}.$$ (Note: Equality case is $a = 51, b = 51, c = 38$.)

We want to find $p_1, p_2, p_3, p_4$ such that \begin{align*} &\sum_{\mathrm{cyc}} \frac{578a^2 + 1143bc}{p_1a^2 + p_2a(b + c) + p_3(b^2 + c^2) + p_4 bc}\\ &\qquad \times\sum_{\mathrm{cyc}} \frac{p_1a^2 + p_2a(b + c) + p_3(b^2 + c^2) + p_4 bc}{527(a + b + c)^2} \\ \le{}& \frac{253^2}{140^2}. \end{align*}

By the equality case, we have $$p_4 = \frac{1073}{578}p_1 - \frac{5479}{3162}p_3 + \frac{1181}{1054}p_2.$$

By the way, there are other forms using Cauchy-Bunyakovsky-Schwarz inequality, e.g. \begin{align*} &\sum_{\mathrm{cyc}} \frac{1}{p_1a^2 + p_2a(b + c) + p_3(b^2 + c^2) + p_4 bc}\\ &\qquad \times\sum_{\mathrm{cyc}} \frac{[(p_1a^2 + p_2a(b + c) + p_3(b^2 + c^2) + p_4 bc)](578a^2 + 1143bc)}{527(a + b + c)^2} \\ \le{}& \frac{253^2}{140^2}. \end{align*}