Notions of logic

[vague bramble]

show that $(a+b)^2 \leq 2$

[quartz crane]

bloubbloub

[vague bramble]

hint: || inegalité arithmetico geometrique ||

[spare gate]

I haven't studied it yet

[vague bramble]

then you can write $(a+b)^2 = 2a^2 + 2b^2 - (a^2 - 2ab + b^2)$

[quartz crane]

bloubbloub

[vague bramble]

see what you can do with this

[spare gate]

No it doesn't work

[vague bramble]

did you factorise the left side

$(a+b)^2 = 2a^2 + 2b^2 - (a-b)^2$

[quartz crane]

bloubbloub

[vague bramble]

and $-(a-b)^2 \leq 0$

[quartz crane]

bloubbloub

[spare gate]

I found it

[spare gate]

.close