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Question Number 157749 by naka3546 last updated on 27/Oct/21
a^2 + b^2 + c^2 + d^2 = 4 a,b,c,d ∈ R max{a^3 + b^3 + c^3 + d^3 } = ?
$${a}^{\mathrm{2}} +\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:+\:{d}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$${a},{b},{c},{d}\:\in\:\mathbb{R} \\ $$$${max}\left\{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:+\:{d}^{\mathrm{3}} \right\}\:\:=\:\:? \\ $$
Commented by naka3546 last updated on 28/Oct/21
Which inequality is used for solving it ?
$$\mathrm{W}{hich}\:\:{inequality}\:\:{is}\:\:{used}\:\:{for}\:\:{solving}\:\:{it}\:? \\ $$
Commented by mr W last updated on 27/Oct/21
max=8
$${max}=\mathrm{8} \\ $$
Commented by naka3546 last updated on 27/Oct/21
Show your workings, please, sir.
$${Show}\:\:{your}\:\:{workings},\:{please},\:{sir}. \\ $$
Commented by naka3546 last updated on 27/Oct/21
Thank you, sir.
$${Thank}\:\:{you},\:{sir}. \\ $$
Commented by mr W last updated on 27/Oct/21
if x_1 +x_2 +x_3 +...+x_n =s with x_i ≥0, then for p>1 (x_1 ^p +x_2 ^p +x_3 ^p +...+x_n ^p )_(min) =n((s/n))^p =(s^p /n^(p−1) ) (x_1 ^p +x_2 ^p +x_3 ^p +...+x_n ^p )_(max) =s^p
$${if}\:{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +…+{x}_{{n}} ={s}\:{with}\:{x}_{{i}} \geqslant\mathrm{0}, \\ $$$${then}\:{for}\:{p}>\mathrm{1} \\ $$$$\:\left({x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +…+{x}_{{n}} ^{{p}} \right)_{{min}} ={n}\left(\frac{{s}}{{n}}\right)^{{p}} =\frac{{s}^{{p}} }{{n}^{{p}−\mathrm{1}} } \\ $$$$\:\left({x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +…+{x}_{{n}} ^{{p}} \right)_{{max}} ={s}^{{p}} \\ $$
Commented by mr W last updated on 03/Aug/23
i don′t know if i′m using a known inequality. actually i just use my brain and logic. let′s look at f(x)=x^p with p>1.
$${i}\:{don}'{t}\:{know}\:{if}\:{i}'{m}\:{using}\:{a}\:{known} \\ $$$${inequality}.\:{actually}\:{i}\:{just}\:{use}\:{my} \\ $$$${brain}\:{and}\:{logic}. \\ $$$${let}'{s}\:{look}\:{at}\:{f}\left({x}\right)={x}^{{p}} \:{with}\:{p}>\mathrm{1}.\: \\ $$
Commented by mr W last updated on 28/Oct/21
Commented by mr W last updated on 03/Aug/23
we can see (certainly it′s also easy to prove): f(x_1 +x_2 )≥f(x_1 )+f(x_2 ), i.e. (x_1 +x_2 )^p ≥x_1 ^p +x_2 ^p (equality holds when x_2 =0) applying this we get (x_1 +x_2 +x_3 )^p ≥(x_1 +x_2 )^p +x_3 ^p ≥x_1 ^p +x_2 ^p +x_3 ^p and ... (x_1 +x_2 +x_3 +...+x_n )^p ≥x_1 ^p +x_2 ^p +x_3 ^p +...+x_n ^p ⇒x_1 ^p +x_2 ^p +x_3 ^p +...+x_n ^p ≤(x_1 +x_2 +x_3 +...+x_n )^p =s^p i.e. (x_1 ^p +x_2 ^p +x_3 ^p +...+x_n ^p )_(max) =s^p
$${we}\:{can}\:{see}\:\left({certainly}\:{it}'{s}\:{also}\:{easy}\right. \\ $$$$\left.{to}\:{prove}\right): \\ $$$${f}\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)\geqslant{f}\left({x}_{\mathrm{1}} \right)+{f}\left({x}_{\mathrm{2}} \right),\:{i}.{e}. \\ $$$$\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{{p}} \geqslant{x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} \\ $$$$\left({equality}\:{holds}\:{when}\:{x}_{\mathrm{2}} =\mathrm{0}\right) \\ $$$${applying}\:{this}\:{we}\:{get} \\ $$$$\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \right)^{{p}} \geqslant\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{{p}} +{x}_{\mathrm{3}} ^{{p}} \geqslant{x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} \\ $$$${and}\:… \\ $$$$\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +…+{x}_{{n}} \right)^{{p}} \geqslant{x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +…+{x}_{{n}} ^{{p}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +…+{x}_{{n}} ^{{p}} \leqslant\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +…+{x}_{{n}} \right)^{{p}} ={s}^{{p}} \\ $$$${i}.{e}. \\ $$$$\left({x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +…+{x}_{{n}} ^{{p}} \right)_{{max}} ={s}^{{p}} \\ $$
Commented by mr W last updated on 28/Oct/21
back to the question, we have x_1 =a^2 , x_2 =b^2 , x_3 =c^2 , x_4 =d^2 s=a^2 +b^2 +c^2 +d^2 =x_1 +x_2 +x_3 +x_4 =4 p=(3/2)>1 a^3 +b^3 +c^3 +d^3 =x_1 ^p +x_2 ^p +x_3 ^p +x_4 ^p ≤4^p =4^(3/2) =8 ⇒(a^3 +b^3 +c^3 +d^3 )_(max) =8
$${back}\:{to}\:{the}\:{question},\:{we}\:{have} \\ $$$${x}_{\mathrm{1}} ={a}^{\mathrm{2}} ,\:{x}_{\mathrm{2}} ={b}^{\mathrm{2}} ,\:{x}_{\mathrm{3}} ={c}^{\mathrm{2}} ,\:{x}_{\mathrm{4}} ={d}^{\mathrm{2}} \\ $$$${s}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} ={x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} =\mathrm{4} \\ $$$${p}=\frac{\mathrm{3}}{\mathrm{2}}>\mathrm{1} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} ={x}_{\mathrm{1}} ^{{p}} +{x}_{\mathrm{2}} ^{{p}} +{x}_{\mathrm{3}} ^{{p}} +{x}_{\mathrm{4}} ^{{p}} \leqslant\mathrm{4}^{{p}} =\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{8} \\ $$$$\Rightarrow\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} \right)_{{max}} =\mathrm{8} \\ $$
Commented by naka3546 last updated on 29/Oct/21
Thank you so much , sir.
$${Thank}\:\:{you}\:\:{so}\:\:{much}\:,\:\:{sir}. \\ $$

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