PREDICTION: Lebron/AD Lakers will 3-peat just like Shaq/Kobe Lakers

[keep-itreal]

And Lebron will tie Jordan with 6 rings:rockon::rockon::rockon:

[Marchesk]

And Lebron will tie Jordan with 6 rings:rockon::rockon::rockon:

Not three, not four, not five ...

[999Guy]

yeah they feel borderline unfair. But like as you said, Kobe/Shaq unfair instead of literally unfair of Curry/KD/Dray/Iggy/Klay.


AD is pretty injury prone but I’ll say this, it would be incredibly difficult to defeat a player of LeBron’s caliber with a resilient MVP level guy next to him. Like, incredibly. Even if you could push this team, when LeBron goes 2016 finals mode in those games 5-7, it would be a monumental task to win.

And usually duos get better as far as play style in that second year. So LA be at the least even sharper offensively next year when the chemistry is peaking and they know all the sets and plays that get stopped the least and they can just go to it in 4th quarters and big games.

[ThatCoolKid]

Only problem is Lebron is getting older. And some people could make another super team in the next few years a la Giannis to the Mavs/Warriors.

[StrongLurk]

0 chance that happens...at least wait until the Lakers win ONE first. ISH man...

[NBAGOAT]

they'll need kuzma to get better with lebron dropping off a bit every year. I also doubt they go hard in the rs again like they did this year. this team wanted a 1 seed but older cores usually dont go hard for multiple years. say like den or gs have a pretty great home court advantage so it's just not easy to 3 peat

[Nike D'Antoni]

KCP, Howard, Rondo, Morris doubt all 4 stays. They will cost ton of money.
They have been huge factors. Unless Buss willing to pay crazy amounts.

[ThatCoolKid]

KCP, Howard, Rondo, Morris doubt all 4 stays. They will cost ton of money.
They have been huge factors. Unless Buss willing to pay crazy amounts.

No one will pay Rondo or Howard.

[Nike D'Antoni]

No one will pay Rondo or Howard.

Title contending teams will. Heat are paying Iguodala 15 million.