Remark. In this topic, I'll give two example to introduce a method called Contradiction on three yields which is represented in synonym tag "$CY3$".
Example 1. Given $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=3.$ Prove that$$\frac{1}{\sqrt{5a+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}}\geq1.$$This problem was discussed here.
Michael Rozenberg gave a proof using the Contradiction method. The main idea is that using contradicted hypothesis makes directly a contradiction to the original given condition.
In here, we also can use this method in "immediate" way. See also here.
"Indeed, by the relation between 3 yields $$x^2y^2+y^2z^2+z^2x^2; x^2y^2z^2; x+y+z,$$ we just need to prove a inequality without radicals."
Proof of example 1.
Firstly, we'll prove the following problem as a lemma.
$(*)$ For all $x,y,z\ge 0$ satisfying$$x^2y^2+y^2z^2+z^2x^2+\frac{33}{16}x^2y^2z^2\ge \frac{81}{16},$$then $$x+y+z\ge 3.$$ Equality holds at $x=y=z=1;(x,y,z)=\left(0;\dfrac{3}{2};\dfrac{3}{2}\right).$
Proof of lemma.
Assuming that $0<x+y+z<3,$ let $m=kx;n=ky;p=kz$ where $k=\dfrac{3}{x+y+z}>1.$
Now, $m>x,n>y,p>z$ and $m+n+p=3.$ We'll prove$$m^2n^2+n^2p^2+p^2m^2+\frac{33}{16}m^2m^2p^2\le \frac{81}{16} (*).$$ Let $p=m+n+p; q=mn+np+pm; r=mnp.$ Rewrite $(*)$ in form $pqr$, a function of $r$$$f(r)=-\frac{33}{16}r^2+6r+\frac{81}{16}-q^2\ge 0,$$where $f'(r)=6-\dfrac{33}{8}r>0$ thus $f(r)$ is increasing.
By Schur of third degree, \begin{align*} f(r)&\ge f\left(\frac{4q-9}{3}\right)\\&=\frac{-33}{16}\left(\frac{4q-9}{3}\right)^2+6\left(\frac{4q-9}{3}\right)+\frac{81}{16}-q^2\\&=\frac{14}{3}(3-q)\left(q-\frac{9}{4}\right)\ge 0,\forall \frac{9}{4}\le q\le 3. \end{align*} In case $0\le q\le \dfrac{9}{4}:$ $f(r)=3r\left(2-\dfrac{11}{16}r\right)+\left(q+\dfrac{9}{4}\right)\left(-q+\dfrac{9}{4}\right)\ge 0.$
Hence, the $(*)$ is proven.
Also by the condition, $m>x,n>y,p>z$ $$m^2n^2+n^2p^2+p^2m^2+\frac{33}{16}m^2m^2p^2>x^2y^2+y^2z^2+z^2x^2+\frac{33}{16}x^2y^2z^2\ge \frac{81}{16} (**).$$ From $(*),(**)$ we get a contradiction.
In conclusion, the fact is proven.
Now, by denoting$$x=\frac{3}{\sqrt{5a+4}};y=\frac{3}{\sqrt{5b+4}};z=\frac{3}{\sqrt{5c+4}},$$which the lemma $(*)$ says that it's enough to prove that $$81\sum_{cyc}\frac{1}{(5a+4)(5b+4)}+\frac{33}{16}.\frac{729}{(5a+4)(5b+4)(5c+4)}\ge \frac{81}{16},$$or ($p=a+b+c; r=abc$) $$16\left(5p+12\right)+297\ge 125r+80p+100q+64 \iff 1\ge abc.$$The last inequality is trivial by AM-GM.
Hence, the proof is done. Equality holds at $a=b=c=1$ or $a=b\rightarrow 0; c\rightarrow +\infty.$
Example 2. Given non-negative real numbers $a,b,c: ab+bc+ca>0.$ Prove that$$\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge \frac{\sqrt{2}}{\sqrt{a+b+c}}.$$Equality holds at: $a=b\rightarrow 0;c\rightarrow +\infty$ or $a=b\rightarrow +\infty;c\rightarrow 0$ or $a=b=t>0;c=0$ or $a=b=t>0;c\rightarrow +\infty$ and permutations.
This problem was discussed here.
Michael Rozenberg gave a proof by Holder using. We also can use the Contradiction method by using the lemma $(*).$
I'll demonstrate the similar idea by optimized way. In the following proof, we'll find out the relation between 2 yields $$x^2y^2+y^2z^2+z^2x^2; x+y+z,$$and the last is proving a inequality without radicals.
Proof of exmaple 2.
WLOG assuming that $a+b+c=2,$ we'll prove that$$\sqrt{\frac{a}{4a^2+bc}}+\sqrt{\frac{b}{4b^2+ac}}+\sqrt{\frac{c}{4c^2+ba}}\ge 1.$$ Now, we use an old problem as a lemma.
$(**)$ For all $x,y,z\ge 0$ then $$(x+y+z)^4\ge 16(x^2y^2+y^2z^2+z^2x^2).$$I proved it by $pqr$ method. We rewrite the inequality as$$p^4\ge 16(q^2-2pr).$$WLOG, assuming $p=2$, we'll prove $4r+1\ge q^2.$
By Schur of third degree, $r\ge \dfrac{8(q-1)}{9},$ we obtain$$1+\dfrac{32(q-1)}{9}-q^2\ge 0 \iff (q-1)\left(q-\frac{23}{9}\right)\le 0, \forall 1\le q\le \frac{4}{3}.$$In the remain case $0\le q\le 1,$ it is trivial.
Hence, the $(**)$ is proven. Equality holds at $(x,y,z)=(0,t,t).$
Now, by denoting$$x=\sqrt{\frac{a}{4a^2+bc}};y=\sqrt{\frac{b}{4b^2+ac}};z=\sqrt{\frac{c}{4c^2+ba}},$$which the lemma $(**)$ says that it's enough to prove that $$16\left[ab(4c^2+ab)+bc(4a^2+bc)+ca(4b^2+ca)\right]\ge (4a^2+bc)(4b^2+ca)(4c^2+ab).$$ It is equivalent to $pqr$ form $$f(r)=-125r^2+r(32+120q)+16q^2-16q^3,$$for $0\le q=ab+bc+ca\le \dfrac{4}{3};0\le r=abc\le \dfrac{8}{27}$.
Notice that $f'(r)=120q-250r+32>0 $ thus $f(r)$ is increasing.
By using Schur of third degree \begin{align*} f(r)&\ge f\left(\frac{8(q-1)}{9}\right)\\&=-125\left(\frac{8(q-1)}{9}\right)^2+\left(\frac{8(q-1)}{9}\right)(32+120q)+16q^2-16q^3\\&=\frac{16}{81}(1-q)(81q^2-40q-644)\ge 0, \forall 1\le q\le \frac{4}{3}. \end{align*} In the remain case $0\le q\le 1$ it is true since$$f(r)=16q^2(1-q)+r(32+120q-125r) \ge 0 .$$Hence, the proof is done.
Some thoughts. In my opinion, the Contradiction on three yields is also just an option which can be appropriate in some cases of some problems.
By using this method, we can eliminate the radicals which Holder or others classical method is hard to use.
I have two opened question.
How can we make it more general ?
In some problems I've worked on, the relation between 3 yields $$x^2y^2+y^2z^2+z^2x^2; x^2y^2z^2; x+y+z,$$ is not enough.
It motivated me make more relations, for example $x^2y^2+y^2z^2+z^2x^2; x^2y^2z^2; x+y+z,x^2+y^2+z^2,...$ The rest is proving an inequality by $uvw$ but it is ugly in calculus.
Is it eligible to accepted as an synonyms tag ? If so, how can I add it is a new tag ?
I'll add updated editings about $CY3$ in this topic for later in case I find some thing interesting.
If you have problems which can apply the Contradiction method, please feel free to post it at comment part.
I really hope to see responses about the topic. All comment and answer is welcome. Thank you for interest.
