Having $c\gt0$
$a, c \in \Bbb{R}$
$$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$$
I don't know how to solve this inequality, I tried with the AM-GM inequality and I reached this point so far.
$$(a+b)^2\le a^2(c+\frac{1}{c})+b^2(c+\frac{1}{c})$$
I don't know how to continue, and I don't even know if this is the right path.
If you have any hints, those would be much appreciated. Thanks
Hi others have given you solutions. I only want to comment on your partial results but since I cannot comment now I'll type it in the answer box.
If you have already proved $(a+b)^2 \leqslant (c+\frac{1}{c}) (a^2+b^2)$ then you are done because $RHS\geqslant 2(a^2+b^2) \geqslant (a+b)^2=LHS$.
However I believe you made a mistake because the following inequality cannot always hold:
$(1+c)a^2 + (1+\frac{1}{c})b^2 \geqslant (c+\frac{1}{c})(a^2+b^2)$.
For example if $c=2$, it reduces to $3a^2+\frac{3}{2}b^2 \geqslant \frac{5}{2} (a^2+b^2) \iff a^2\geqslant 2b^2$ so if $a=1, b=1$ it's reversed.
We have that
$$(a+b)^2\le(1+c)a^2+\left(1+\frac{1}{c}\right)b^2 $$
$$a^2+b^2+2ab\le a^2+b^2+ca^2+\frac1cb^2$$
$$2ab\le ca^2+\frac1cb^2 \iff \frac{ca^2+\frac1cb^2}2 \ge ab$$
which is true by AM-GM.
hint
Put $$c=d^2, A=ad \;and\; B=\frac bd$$
By AM-GM, we have
$$2AB\le A^2+B^2$$
or
$$2ab\le a^2c+\frac{b^2}{c}$$
By C-S $$(1+c)a^2+\left(1+\frac{1}{c}\right)b^2=$$ $$=\left(\frac{1}{1+c}+\frac{c}{1+c}\right)\left((1+c)a^2+1+\frac{(1+c)b^2}{c}\right)\geq(a+b)^2.$$
