$(a+b)^2\le(1+c)a^2+(1+\frac{1}{c})b^2$ Proof

Question

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

Answer 1

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.

Answer 2

By C-S $$(1+c)a^2+\left(1+\frac{1}{c}\right)b^2=$$ $$=\left(\frac{1}{1+c}+\frac{c}{1+c}\right)\left((1+c)a^2+1+\frac{(1+c)b^2}{c}\right)\geq(a+b)^2.$$

Answer 3

hint

Put $$c=d^2, A=ad \;and\; B=\frac bd$$

By AM-GM, we have

$$2AB\le A^2+B^2$$

or

$$2ab\le a^2c+\frac{b^2}{c}$$

Answer 4

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.