# Convicted of squares in an Inequality and my Weakness

It had been 3 and half months I'm sitting and revolving around Inequalities, but I am still not able to properly grasp even the most fundamental Inequalities like Cauchy's Inequality and the AM-GM Inequality. I've tried books of Pham Kim Hung, Zdravko Cvetkovski and the book Inequalities - A Mathematical Olympiad Approach, but none are helpful. Any good recommendations or help that gets me on the right track and because of which my time becomes more fruitful would be considered heavenly.
These problems are credited to Samin Riasat's Basics of Olympiad Inequalities( yet another book! ) and for a note that these inequalities are to be solved by methods that are seriously elementary. In fact, only Cauchy's Inequality and the AM-GM Inequality are to be employed, and nothing beyond. These can be taken as examples to explain me what intuition and knowledge are essential for proving Inequalities that are on a step ahead of the basics-

$$1.$$ Let a, b, c be positive real numbers such that $$a + b + c = 1$$. Prove that $$\frac{a}{\sqrt {a+2b}}+\frac{b}{\sqrt {b+2c}}+\frac{c}{\sqrt {c+2a}} \lt \sqrt{\frac{3}{2}}$$ Here I provide a little space for you to understand what I don't-

I used Cauchy as till this point in the book only Inequalities taught were Cauchy's Inequality and the AM-GM Inequality.
I first transformed the structure of the proposition into the standard form of C-S as follows $$\left(\frac{a}{\sqrt {a+2b}}+\frac{b}{\sqrt {b+2c}}+\frac{c}{\sqrt {c+2a}}\right)^2 \lt \frac{3}{2}$$ And then removed the '$$\lt \frac{3}{2}$$' for a while to get a feel of the LHS. On the next step, I remembered as C-S by the intuition that when the lesser side is given, it must be the summation of a product of two quantities that are to be separated in each term and squared, summed separately by squaring and summing all the first factors and the second factors and finally multiplied.

The innovation is to be applied now, here in this step.
A natural question is that what two factors are the terms to be broken into? This is the step where I require advice.
My attempt was this- $$\left(\sqrt{a} \times \frac{\sqrt{a}}{\sqrt {a+2b}}+\sqrt{b} \times \frac{\sqrt{b}}{\sqrt {b+2c}}+\sqrt{b} \times \frac{\sqrt{c}}{\sqrt {c+2a}}\right)^2 \lt \frac{3}{2}$$ LHS is
$$\le \left(a+b+c\right)\left(\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2b}\right)=\left(\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2b}\right)$$ by the constraint in the question. But it leads to- $$\frac{a}{a+2b}+\frac{b}{b+2c}+\frac{c}{c+2b} \lt \frac{3}{2}$$ And here my attempt fails. I don't know from which hell did that strict Inequality came from, and how to prove the remainder of my attempt. I don't know if true or false, but feel this approach was to simple to obliterate the problem and some more wilderness is required.
I wish to know whether my choice of books are too advanced, or the questions are too difficult or something else that objects my progress?
Finally here is another question to which I need a solution-

$$2.$$Let $$a, b, c > 0$$. Prove that $$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}}\le \sqrt{3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$$ The same approach was applied but unsuccessful.

• Is it overkill to use Lagrange multipliers? Jul 20, 2020 at 16:04
• Lagranage multipliers are very advanced for me to use. (As a beginner) Jul 20, 2020 at 16:30

By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{a+2b}}\right)^2\leq\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\sum_{cyc}a(a+2c)=\sum_{cyc}\frac{a}{(a+2b)(a+2c)}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{a}{(a+2b)(a+2c)}\leq\frac{3}{2(a+b+c)},$$ which is obvious after full expanding: $$\sum_{sym}(2a^4bc+2a^3b^3+24a^3b^2c+12.5a^2b^2c^2)\geq0.$$ A proof of the second problem see here: Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$
• @Book Of Flames I used Cauchy-Schwarz inequality. $(a_1b_1+a_2b_2+...+a_nb_n)^2\leq(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n)^2.$ Jul 20, 2020 at 16:40
• @Book Of Flames I did not know before that it will give a proof. I tried to get a symmetric expression. Also $\sum\limits_{cyc}(a^2+2ac)=1$. It's just turned out. Just prove inequalities and you'll get similar solutions by yourself. Jul 20, 2020 at 17:05