Given three positive numbers $a, b, c$ so that $a\leqq b\leqq c$. Prove that $$\sum\limits_{cyc}\,\frac{a+ 1.4\,b}{1.4\,a+ b}\geqq 3$$
$$\because\sum\limits_{cyc}\frac{a+ b}{7a+ 5b}\geqq \frac{1}{2}$$ $$\because\sum\limits_{cyc}\left ( (a+ b)(7b+ 5c)(7c+ 5a) \right )\geqq \frac{1}{2}\prod\limits_{cyc}(7a+ 5b)$$ $$\because 95\sum\limits_{cyc}ab^{2}+ 119\sum\limits_{cyc}a^{2}b+ 222abc\geqq \frac{1}{2}\left ( 95\sum\limits_{cyc}ab^{2}+ 245\sum\limits_{cyc}a^{2}b+ 468abc \right )$$ $$\because 15\sum\limits_{cyc}ab^{2}- 7\sum\limits_{cyc}a^{2}b- 24abc\geqq 0$$ $$\because 15\left ( \sum\limits_{cyc}(b- c)a^{2} \right )+ 8(\sum\limits_{cyc}a^{2}b- 3abc)\geqq 0$$ $$\because 15\sum\limits_{cyc}(b- c)a^{2}\geqq 0$$ We will prove $\sum\limits_{cyc}(b- c)a^{2}\geqq 0$ so that $0< a\leqq b\leqq c$ involving to Jensen inequality, like above : $$(b- c)a^{2}+ (a- b)c^{2}\geqq (b- c+ a- b)\left ( \frac{(b- c)a+ (a- b)c}{b- c+ a- b} \right )^{2}$$ Is true? (So how about another solutions here!?)
For $K= \dfrac{7}{5}$ it's wrong. Try $(a,b,c)= (25,9,1).$
For $a\leq b\leq c$ we need to prove that $$\sum_{cyc}(15a^2c-7a^2b-8abc)\geq 0$$ or $$8(a^2c+b^2a+c^2b-3abc)+7(a-b)(b-c)(c-a)\geq 0,$$ which is true by AM-GM.