Let $x+y+z=1$, $x,y,z> 0$. Show that $\sum_{cyc}\sqrt{1-2xy}\geq\sqrt7$ by Jensen's inequality?

Question

Let $x+y+z=1$, $x,y,z> 0$. Show that $\sum_{cyc}\sqrt{1-2xy}\geq\sqrt7.$

Calculus way:

Let $A=\sqrt{1-2xy}$, $B=\sqrt{1-2yz}$, $C=\sqrt{1-2zx}$, $f(x,y,z)=A+B+C$ and $g(x,y,z)=x+y+z-1.$ By the method of Lagrange multipliers ($f_{x_i}=\lambda g_{x_i}$), we can find the system $$cyc: \frac{y-x}{z}=\frac{AC-AB}{BC}$$ and $$cyc: x(y-x)B=z(y-z)A.$$ If we multiply all equation in the last system and cancel $xyzABC$ we find $$(y-x)(z-y)(x-z)=(y-z)(z-x)(x-y)$$ and $x=y=z$ as required.

I tried but I couldn't complete my solution with weighted Jansen's inequality. Can you suggest another method? Thanks in advance.

Answer 1

By Holder $$\sum_{cyc}\sqrt{1-2ab}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{1-2ab}\right)^2\sum\limits_{cyc}\frac{(2c+3)^3}{1-2ab}}{\sum\limits_{cyc}\frac{(2c+3)^3}{1-2ab}}}\geq\sqrt{\frac{1331}{\sum\limits_{cyc}\frac{(2c+3)^3}{1-2ab}}}$$ and after homogenization it's enough to prove that: $$1331(a+b+c)\geq7\sum_{cyc}\frac{(3a+3b+5c)^3}{(a+b+c)^2-2ab}$$ and the rest is smooth: a full expanding gives $$\sum_{sym}(19.5a^6+342a^6b+492a^5b^2+39a^4b^3+219.5a^5bc+414a^4b^2c+205a^3b^3c-1731a^3b^2c^2)\geq0,$$ which is true by Muirhead.

Answer 2

Another way.

We need to prove that: $$\sum_{cyc}\sqrt{a^2+b^2+c^2+2ab+2ac}\geq\sqrt7(a+b+c)$$ or $$\sum_{cyc}\left(3a^2+4ab+2\sqrt{(a^2+b^2+c^2+2ab+2ac)(a^2+b^2+c^2+2ab+2bc)}\right)\geq7\sum_{cyc}(a^2+2ab)$$ or $$\sum_{cyc}(-4a^2-10ab+6a^2+8ab)\geq\left(\sqrt{a^2+b^2+c^2+2ab+2ac}-\sqrt{a^2+b^2+c^2+2ab+2bc}\right)^2$$ or $$\sum_{cyc}(a-b)^2\left(1-\frac{4c^2}{\left(\sqrt{a^2+b^2+c^2+2ab+2ac}+\sqrt{a^2+b^2+c^2+2ab+2bc}\right)^2}\right)\geq0,$$ which is true because $$\sqrt{a^2+b^2+c^2+2ab+2ac}+\sqrt{a^2+b^2+c^2+2ab+2bc}\geq c+c=2c.$$

Answer 3

Another way.

Let $c=\min\{a,b,c\}$ and $f(a,b,c)=\sum\limits_{cyc}\sqrt{1-2ab}.$

Thus, $c\leq\frac{1}{3}$ and $$f(a,b,c)-f\left(\frac{a+b}{2},\frac{a+b}{2},c\right)=$$ $$=\sqrt{1-2ab}-\sqrt{1-\frac{(a+b)^2}{2}}+\sqrt{1-2ac}+\sqrt{1-2bc}-2\sqrt{1-c(a+b)}=$$ $$=\tfrac{\frac{(a+b)^2}{2}-2ab}{\sqrt{1-2ab}+\sqrt{1-\frac{(a+b)^2}{2}}}+\tfrac{1-2ac+1-2bc+2\sqrt{(1-2ac)(1-2bc)}-4+4c(a+b)}{\sqrt{1-2ac}+\sqrt{1-2bc}+2\sqrt{1-c(a+b)}}=$$ $$=\tfrac{(a-b)^2}{2\left(\sqrt{1-2ab}+\sqrt{1-\frac{(a+b)^2}{2}}\right)}-\tfrac{\left(\sqrt{1-2ac}-\sqrt{1-2bc}\right)^2}{\sqrt{1-2ac}+\sqrt{1-2bc}+2\sqrt{1-c(a+b)}}=$$ $$=(a-b)^2\left(\tfrac{1}{2\left(\sqrt{1-2ab}+\sqrt{1-\frac{(a+b)^2}{2}}\right)}-\tfrac{4c^2}{\left(\sqrt{1-2ac}+\sqrt{1-2bc}+2\sqrt{1-c(a+b)}\right)\left(\sqrt{1-2ac}+\sqrt{1-2bc}\right)^2}\right)\geq$$ $$\geq(a-b)^2\left(\tfrac{1}{4}-\tfrac{4\cdot\frac{1}{9}}{\left(2\sqrt{a^2+b^2+c^2}+2\sqrt{1-c(1-c)}\right)\left(2\sqrt{a^2+b^2+c^2}\right)^2}\right)\geq$$ $$\geq(a-b)^2\left(\tfrac{1}{4}-\tfrac{4}{9(1+2\sqrt{1-c+c^2})}\right)=(a-b)^2\left(\tfrac{1}{4}-\tfrac{4}{9(1+2\sqrt{\left(c-\frac{1}{2}\right)^2+\frac{3}{4}})}\right)\geq$$ $$\geq(a-b)^2\left(\tfrac{1}{4}-\tfrac{4}{9(1+\sqrt3}\right)=\frac{(a-b)^2(9\sqrt3-7)}{36(1+\sqrt3)}\geq0.$$ Id est, it's enough to prove that $$f\left(\frac{a+b}{2},\frac{a+b}{2},c\right)\geq\sqrt7,$$ which says that it's enough to prove our inequality for equality case of two variables.

Let $b=a$ and $c=1-2a$, where $0\leq a\leq\frac{1}{2}$(we know that we can take even $\frac{1}{3}\leq a\leq\frac{1}{3}$).

Thus, we need to prove that: $$\sqrt{1-2a^2}+2\sqrt{1-2a(1-2a)}\geq\sqrt7$$ or $$2\sqrt{(1-2a^2)(1-2a+4a^2)}\geq1+4a-7a^2$$ and since $$1+4a-7a^2=1+a(4-7a)>0,$$ it remains to prove that: $$4(1-2a^2)(1-2a+4a^2)\geq(1+4a-7a^2)^2$$ or $$(1-3a)^2(3+2a-9a^2)\geq0,$$ which is true because $$3+2a-9a^2\geq3+\frac{2}{3}-\frac{9}{4}>0.$$

Answer 4

We first prove that $$ 63(x+y+z)^{2}((x+y+z)^{2}-2yz)-(7(x^2+y^2+z^2)+17x(y+z)+8yz)^{2}=(\frac{y+z}{2}-x)^{2}(14x^2+28x(y+z)+5(y+z)^{2})+3(\frac{y-z}{2} )^{2}(14x^2+16x(y+z)+17y^2+46yz+17z^2)\ge 0 $$ then,we have $$ \sum\sqrt{(x+y+z)^2-2yz} \ge \sum \frac{7(x^2+y^2+z^2)+17x(y+z)+8yz}{3\sqrt{7}(x+y+z)}=\sqrt{7}(x+y+z) $$ We have completed the proof.