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Let-a-b-c-gt-0-42abc-4bca-1cab-24-Then-1-ab-2-bc-3-ca-Z-A-




Question Number 164856 by Zaynal last updated on 22/Jan/22
    Let a,b,c >0;      42abc + 4bca +1cab ≤ 24      Then (1/(ab)) + (2/(bc)) + (3/(ca)) = ??                  ^({Z.A})
$$\:\:\:\:\boldsymbol{\mathrm{Let}}\:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:>\mathrm{0}; \\ $$$$\:\:\:\:\mathrm{42}\boldsymbol{{abc}}\:+\:\mathrm{4}\boldsymbol{{bca}}\:+\mathrm{1}\boldsymbol{{cab}}\:\leqslant\:\mathrm{24} \\ $$$$\:\:\:\:\mathrm{T}\boldsymbol{\mathrm{hen}}\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{ab}}}\:+\:\frac{\mathrm{2}}{\boldsymbol{\mathrm{bc}}}\:+\:\frac{\mathrm{3}}{\boldsymbol{\mathrm{ca}}}\:=\:?? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:^{\left\{{Z}.\mathrm{A}\right\}} \\ $$
Commented by MJS_new last updated on 24/Jan/22
abc=bca=cab  anyway you cannot solve this equation with  only one inequation given
$${abc}={bca}={cab} \\ $$$$\mathrm{anyway}\:\mathrm{you}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{equation}\:\mathrm{with} \\ $$$$\mathrm{only}\:{one}\:\mathrm{inequation}\:\mathrm{given} \\ $$

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