Asymmetric, Unusually Restricted Inequality

Question

I don't understand where and how to use the constraint and how to rearrange the inequality to a symmetric form-

Let $a, b, c$ be positive real numbers such that $a \le b \le c$ and $a + b + c = 3$. Prove that $$ \sqrt{3a^2+1} + \sqrt{5a^2+3b^2+1} + \sqrt{7a^2+5b^2+3c^2+1} \le 9 $$

Applying AM-GM Inequality merely gives- $$ 15a^2+8b^2+3c^2 \le 12 $$

Or using C-S- $$ {1\over \sqrt{x}}\sqrt{x(3a^2+1)}+ {1\over \sqrt{y}}\sqrt{y(5a^2+3b^2+1)}+ {1\over \sqrt{z}}\sqrt{z(7a^2+5b^2+3c^2+1)} \le $$ $$ \left({1\over x}+{1\over y}+{1\over z}\right)((3x+5y+7z)a^2+(3y+5z)b^2+(3z)c^2+x+y+z)\le 9 $$ The last line has to be proved, but I don't know how and for which $x,y,z$.

Any help will be thankfully appreciated.

Answer 1

My old solution.

Let $3a^2+1=4x^2,$ $5a^2+3b^2+1=9y^2$ and $7a^2+5b^2+3c^2+1=16z^2,$

where $x,$ $y$ and $z$ are positives.

Thus, by C-S $$\left(\sqrt {3a^2+1} + \sqrt {5a^2+3b^2+1} + \sqrt {7a^2+5b^2+3c^2+1}\right)^2=$$ $$=(2x+3y+4z)^2\leq(2+3+4)(2x^2+3y^2+4z^2)=$$ $$=18x^2+27y^2+36z^2=$$ $$=\frac{9(3a^2+1)}{2}+3(5a^2+3b^2+1)+\frac{9(7a^2+5b^2+3c^2+1)}{4}=$$ $$=\frac{177a^2+81b^2+27c^2+39}{4}.$$ But $$\frac{177a^2+81b^2+27c^2+39}{4}\leq81$$ it's $$531a^2+243b^2+81c^2\leq855$$ or $$531a^2+243b^2+81c^2\leq95(a+b+c)^2$$ or $$14c^2+190bc+190ac+190ab\geq148b^2+436a^2,$$ which is obvious.

Answer 2

Another way.

$$\sqrt{3a^2+1}+\sqrt{5a^2+3b^2+1}+\sqrt{7a^2+5b^2+3c^2+1}\leq$$ $$\leq\left(a+\frac{2b}{3}+\frac{c}{3}\right)+\left(a+\frac{5b}{3}+\frac{c}{3}\right)+(a+b+2c)\leq3(a+b+c)=9.$$ Because $$\sqrt{3a^2+1}\leq a+\frac{2b}{3}+\frac{c}{3}$$ it's $$27a^2+(a+b+c)^2\leq(3a+2b+c)^2$$ or $$27a^2+a^2+b^2+c^2+2ab+2ac+2bc\leq9a^2+4b^2+c^2+12ab+6ac+4bc$$ or $$19a^2\leq3b^2+10ab+4ac+2bc,$$ which is true for $0<a\leq b\leq c$.

Also, $$\sqrt{5a^2+3b^2+1}\leq a+\frac{5b}{3}+\frac{c}{3}$$ it's $$45a^2+27b^2+(a+b+c)^2\leq(3a+5b+c)^2$$ or $$45a^2+27b^2+a^2+b^2+c^2+2ab+2ac+2bc\leq9a^2+25b^2+c^2+30ab+6ac+10bc$$ or $$37a^2+3b^2\leq28ab+4ac+8bc,$$ which is obvious again.

A proof of $$\sqrt{7a^2+5b^2+3c^2+1}\leq a+b+2c$$ is the same.